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You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.


Solution:
(a) \(\frac{\pi}{6}\)
Hint:

Question 2.


Solution:
(c) \(\frac{5}{2}\)
Hint:

Question 3.


Solution:
(c) 0
Hint:

Question 4.


Solution:
(d) \(\frac{2}{3}\)
Hint:
It is an even function

Question 5.

(b) 2π
(c) 3π
(d) 4π
Solution:
(d) 4π
Hint:

Question 6.

(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:

Question 7.

(a) cos x – x sin x
(b) sin x + x cos x
(c) x cos x
(d) x sin x
Solution:
(c) x cos x
Hint:

Question 8.
The area between y² = 4x and its latus rectum is ………

Solution:
(c) \(\frac{8}{3}\)
Hint:

Question 9.


Solution:
(b) \(\frac{1}{10100}\)

Question 10.


Solution:
(a) \(\frac{\pi}{2}\)

Question 11.

(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9
Hint:

Question 12.


Solution:
(b) \(\frac{2}{9}\)
Hint:

Question 13.


Solution:
\(\frac{3 \pi}{8}\)
Hint:

Question 14.


Solution:
(d) \(\frac{2}{27}\)
Hint:

Question 15.

(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:

Question 16.
The volume of solid of revolution of the region bounded by y² = x(a – x) about x-axis is ……..

Solution:
(d) \(\frac{\pi a^{3}}{6}\)
Hint:

Question 17.

(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:

Question 18.


Solution:
(d) \(\frac{\pi^{2}}{4}-2\)
Hint:

Question 19.


Solution:
(b) \(\frac{3 \pi a^{4}}{16}\)
Hint:

Question 20.


Solution:
(a) \(\frac{1}{2}\)
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The area bounded by the line y = x, the x – axis, the ordinates x = 1,x = 2 is …….

Solution:
(a) \(\frac{3}{2}\)
Hint:

Question 2.
The area of the region bounded by the graph of y = sin x and y = cos x between x = 0 and x = \(\frac{\pi}{4}\) is ……..

Solution:
(b) \(\sqrt{2}-1\)
Hint:

Question 3.
The area between the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and its auxiliary circle is …….
(a) πb(a – b)
(b) 2πa(a – b)
(c) πa(a – b)
(d) 2πb(a – b)
Solution:
(c) πa(a – b)
Hint:

Question 4.
The area bounded by the parabola y² = x and its latus rectum is ……..

Solution:
(b) \(\frac{1}{6}\)
Hint:

Question 5.
The volume of the solid obtained by revolving \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 about the minor axis is …….
(a) 48π
(b) 64π
(c) 32π
(d) 128π
Solution:
(b) 64π
Hint:

Question 6.
The volume, when the curve y = \(\sqrt{3+x^{2}}\) from x = 0 to x = 4 is rotated about x – axis is ……

Solution:
(c) \(\frac{100}{3} \pi\)
Hint:

Question 7.
The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y – axis is ……….

Solution:
(c) \(\frac{\pi}{3}\)
Hint:

Question 8.
Volume of solid obtained by revolving the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 about major and minor axes are in the ratio …….
(a) b² : a²
(b) a² : b²
(c) a : b
(d) b : a
Solution:
(d) b : a
Hint:

Question 9.
The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) about x-axis is …….
(a) 18π
(b) 2π
(c) 36π
(d) 9π
Solution:
(d) 9π
Hint:

Question 10.
The length of the arc of the curve is …….
(a) 48
(b) 24
(c) 12
(d) 96
Solution:
(a) 48
Hint:
Length of the arc of the curve = 6a

∴ Required length = 6a = 6 × 8 = 48 units.

Question 11.
The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is ……

Solution:
(a) \(8 \sqrt{5} \pi\)
Hint:

Question 12.
The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from the centre is ……
(a) 20π
(b) 40π
(c) 10π
(d) 30π
Solution:
(a) 20π
Hint:
The curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b – a)
Given radius, r = 5; a = 2; b = 4
Required surface area = 2πr (b – a)
= 2π × 5 × (4 – 2) = 20π sq. units

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.6

Question 1.
Find a parametric form of vector equation of a plane which is at a distance of 7 units from the origin having 3, -4, 5 as direction ratios of a normal to it.
Solution:
Given p = 7

Question 2.
Find the direction cosines of the normal to the plane 12x + 3y – 4z = 65. Also, find the non-parametric form of vector equation of a plane and the length of the perpendicular to the plane from the origin.
Solution:
12x + 3y – 4z = 65

(iii) Length of the perpendicular to the plane from the origin is 5 units.

Question 3.
Find the vector and Cartesian equation of the plane passing through the point with position vector \(2 \hat{i}+6 \hat{j}+3 \hat{k}\) and normal to the vector \(\hat{i}+3 \hat{j}+5 \hat{k}\)
Solution:

Question 4.
A plane passes through the point (-1, 1, 2) and the normal to the plane of magnitude \(3 \sqrt{3}\) makes equal acute angles with the coordinate axes. Find the equation of the plane.
Solution:
Given magnitude = \(3 \sqrt{3}\) and \(\vec{a}=-\vec{i}+\vec{j}+2 \vec{k}\)
Then, the normal vector makes equal acute angle with the coordinate axes.
We know that cos² α + cos² β + cos² γ = 1 (But α = β = γ)

Question 5.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})\) = 12 on the coordinate axes.
Solution:

x-intercept = 2; y-intercept = 3; z-intercept = -4

Question 6.
If a plane meets the coordinate axes at A, B, C such that the centroid of the triangle ABC is the point (u, v, w), find the equation of the plane.
Solution:
Let A (a, 0, 0), B(0, b, 0), C(0, 0, c)
centroid of ∆ABC = \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.6 Additional Problems

Question 1.
Find the vector and cartesian equations of a plane which is at a distance of 18 units from the origin and which is normal to the vector \(2 \vec{i}+7 \vec{j}+8 \vec{k}\)
Solution:

Question 2.
Find the unit vector to the plane 2x – y + 2z = 5.
Solution:
Writing the plane in normal form we get,

Question 3.
Find the length of the perpendicular from the origin to the plane \(\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})\) = 26.
Solution:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})\) = 26
i.e., 3x + 4y+ 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is
\(\pm \frac{-26}{\sqrt{9+16+144}}=\frac{+26}{13}\) = 2 units

Question 4.
The foot of the perpendicular drawn from the origin to a plane is (8, -4, 3). Find the equation of the plane.
Solution:
The required plane passes through the point A(8, -4, 3) and is perpendicular to \(\overrightarrow{\mathrm{OA}}\) .

The cartesian equation is 8x – 4y + 3z = 89.

Question 5.
Find the equation of the plane through the point whose position vector is \(2 \vec{i}-\vec{j}+\vec{k}\) and perpendicular to the vector \(4 \vec{i}+2 \vec{j}-3 \vec{k}\).
Solution:
The required plane is perpendicular to \(4 \vec{i}+2 \vec{j}-3 \vec{k}\)
So, it is parallel to the plane 4x + 2y – 3z = k
∴ the equation of the plane is 4x + 2y – 3z = k
The plane passes through the point (2, -1, 1)
⇒ (4)(2) + 2(-1) – 3(1) = λ i.e. λ = 8 – 2 – 3 = 3
So, the equation of the plane is 4x + 2y – 3z = 3.

Question 6.
Find the vector and cartesian equations of the plane passing through the point (2, -1, 4) and parallel to the plane \(\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7.
Solution:
The given plane is \(\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7
i e. \((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7
i.e. 4x – 12y – 3z = 1
The required plane is parallel to the above plane. So, the equation of the required plane is 4x – 12y – 3z – k. The plane passes through (2, -1, 4).
⇒ 4(2) – 12(-1) – 3(4) = k i.e. k = 8 + 12 – 12 = 8
So, the equation of the plane is 4x – 12y – 3z = 8.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5

Question 1.
Find the parametric form of vector equation and Cartesian equations of a straight line passing through (5, 2, 8) and is perpendicular to the straight lines

Solution:

∴ This’ vector is perpendicular to both the given straight lines.
∴ The required straight line is
\(\vec{r}=\vec{a}+t(\vec{b} \times \vec{d})\)

Question 2.
Show that the lines are skew lines and hence find the shortest distance between them.
Solution:

Question 3.
If the two lines intersect at a point, find the value of m.
Solution:

Question 4.
Show that the lines \(\frac{x-3}{3}=\frac{y-3}{-1}\), z – 1 = 0 and \(\frac{x-6}{2}=\frac{z-1}{3}\), y – 2 = 0 intersect. Also find the point of intersection
Solution:

Any point on the Second line

∴ The required point of intersection is (6, 2, 1)

Question 5.
Show that the straight lines x + 1 = 2y = -12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them.
Solution:

Question 6.
Find the parametric form of vector equation of the straight line passing through (-1, 2, 1) and parallel to the straight line \(\vec{r}=(2 \hat{i}+3 \hat{j}-\hat{k})+t(\hat{i}-2 \hat{j}+\hat{k})\) and hence find the shortest distance between the lines.
Solution:

Question 7.
Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\). Also, find the equation of the perpendicular.
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5 Additional Problems

Question 1.
Find the shortest distance between the parallel line

Solution:

Question 2.
Show that the two lines \(\vec{r}=(\vec{i}-\vec{j})+t(2 \vec{i}+\vec{k})\) and \(\vec{r}=(2 \vec{i}-\vec{j})+s(\vec{i}+\vec{j}-\vec{k})\) skew lines and find the distance between them.
Solution:

Question 3.
Show that the lines intersect and hence find the point of intersection
Solution:

Any point on this line is of the form (2µ + 4, 0, 3µ – 1)
Since they are intersecting, for some λ, µ
(3λ + 1, – λ + 1,- 1) = (2µ + 4, 0, 3µ – 1) ⇒ λ = 1 and µ = 0
To find the point of intersection either take λ = 1 or µ = 0
∴ The point of intersection is (4, 0, – 1),

Question 4.
Find the shortest distance between the skew lines.

Solution:
Compare the given equation with \(\vec{r}=\vec{a}_{1}+t \vec{u}\) and \(\vec{r}=\vec{a}_{2}+s \vec{v}\)

Question 5.
Find the shortest distance between the parallel lines.

Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = 2x², y = 0 and x = 1.
Solution:

Question 2.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = e^-2x y = 0, x = 0 and x = 1.
Solution:

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y-axis, the region enclosed by x² = 1 + y and y = 3.
Solution:

Question 4.
The region enclosed between the graphs of y = x and y = x² is denoted by R, Find the volume generated when R is rotated through 360° about x – axis.
Solution:
To find points of intersection, solving y = x² and y = x, we get (0, 0) and (1, 1)

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.
Solution:

Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major-axis 20 cm and minor-axis 10 cm about its major-axis. Find its volume using integration.
Solution:
From the given data a = 10 cm and b = 5 cm
Equation of the Ellipse

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9 Additional Questions

Question 1.
Find the volume of the solid that results when the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (a > b > 0) is revolved about the minor axis.
Solution:
Volume of the solid is obtained by revolving the right side of the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) about the y-axis.
Limits for y is obtained by putting x = 0 ⇒ y² = b² ⇒ y = ±b.

Question 2.
Find the volume of the solid generated when the region enclosed by y = \(\sqrt{x}\), y = 2 and x = 0 is revolved about the y – axis.
Solution:
Since the solid is generated by revolving about the y-axis, rewrite y = \(\sqrt{x}\) as x = y².
Taking the limits for y, y = 0 and y = 2 (Putting x = 0 in x = y², we get y = 0)

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4

Question 1.
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector \(4 \hat{i}+3 \hat{j}-7 \hat{k}\) and parallel to the vector \(2 \hat{i}-6 \hat{j}+7 \hat{k}\)
Solution:

Question 2.
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (-2, 3, 4) and parallel to the straight line \(\frac{x-1}{-4}=\frac{y+3}{5}=\frac{8-z}{6}\)
Solution:

Question 3.
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
Solution:
Given straight line passing through the points (6, 7, 4) and (8, 4, 9).
Direction ratio of the straight line joining these two points 2, -3, -5.
Cartesian equation:

(ii) The straight Line cuts yz-plane
So we get x = 0
2t + 6 = 0 ⇒ 2t = -6
t = -3
-3t + 7 = -3 (-3) + 7 = 9 + 7 = 16
5t + 4 = 5(-3) + 4 = -15 + 4 = -11
The required point (0, 16, -11).

Question 4.
Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7,9,13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Solution:
Given straight line passing through the points (5, 6, 7) and (7, 9, 13)
∴ d.r.s : 2, 3, 6

Note: Selection of \(\vec{a}\) and \(\vec{b}\) is your choice.

Question 5.
Find the acute angle between the following lines.

(iii) 2x = 3y = -z and 6x = -y = -4z
Solution:

Question 6.
The vertices of ∆ABC are A(7, 2, 1), B(6, 0, 3), and C(4, 2, 4) . Find ∠ABC.
Solution:

Question 7.
If the straight line joining the points (2, 1, 4) and (a – 1, 4, -1) is parallel to the line joining the points (0, 2, b – 1) and (5, 3, -2), find the values of a and b.
Solution:

Question 8.
If the straight lines are perpendicular to each other, find the value of m.
Solution:

Question 9.
Show that the points (2, 3, 4),(-1, 4, 5) and (8, 1, 2) are collinear.
Solution:
Given points are (2, 3, 4), (-1, 4, 5) and (8, 1, 2) Equation of the line joining of the first and second point is
\(\frac{x-2}{-3}=\frac{y-3}{1}=\frac{z-4}{1}\) = m (say)
(-3m + 2, m + 3, m + 4)
On putting m = -2, we get the third point is (8, 1, 2)
∴ Given points are collinear.

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 Additional Problems

Question 1.
Find the vector and cartesian equations of the straight line passing through the point A with position vector \(3 \vec{i}-\vec{j}+4 \vec{k}\) and parallel to the vector \(-5 \vec{i}+7 \vec{j}+3 \vec{k}\).
Solution:
We know that vector equation of the line through the point with position vector \(\vec{a}\) and parallel to \(\vec{v}\) is given by \(\vec{r}=\vec{a}+t \vec{v}\) where t is a scalar.
Here \(\vec{a}=3 \vec{i}-\vec{j}+4 \vec{k}\) and \(\vec{v}=-5 \vec{i}+7 \vec{j}+3 \vec{k}\)
Vector equation of the line is
\(\vec{r}=(3 \vec{i}-\vec{j}+4 \vec{k})+t(-5 \vec{i}+7 \vec{j}+3 \vec{k})\) ………………. (1)
The cartesian equation of the line passing through (xp yx, zx) and parallel to a vector whose d.r.s are l, m, n is

Question 2.
Find the vector and cartesian equations of the straight line passing through the points (- 5, 2, 3) and (4, – 3, 6).
Solution:
Vector equation of the straight line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is given by

Question 3.
Find the angle between the lines.

Solution:
Let the given lines be in the direction of \(\vec{u}\) and \(\vec{v}\)

Question 4.
Find the angle between the following lines
Solution:
Angle between two lines is the same as angle between their parallel vectors.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x-axis.
Solution:

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = – 1, y = 3 and y – axis.
Solution:
2x – y + 1 = 0

To find further limit put x = 0, we get y = 1

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x – axis, x = – 3 and x = 3.
Solution:

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
To find point of intersection of the curves
y = 2x + 5 and y = x² – 2x we get (-1, 3) and (5, 15)

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π.
Solution:
To find the points of intersection,

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the line x = 0, x = \(\frac{\pi}{2}\), 0
Solution:
To find the points of intersection of these two curves between 0 to \(\frac{\pi}{2}\) is \(\frac{\pi}{4}\)

Question 7.
Find the area of the region bounded by parabola y² = x and the line y = x – 2
Solution:
To find the points of intersection solve the two equations y² = x and y = x – 2

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4 , y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
To find the points of intersection of the two curves, y² = 4x and x² = 4y are (0, 0) and (4, 4).
Area of the square field = 4 × 4 = 16 sq. units

So, the remaining each of the two parts must be \(\frac{16}{3}\) sq.units.
∴ Yes, It is possible to divide the square field into three equal parts.

Question 9.
The curves = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
y = (x – 2)² + 1



∴ x = 2 is a minimum point
∴ The point P is (2, 1)
But slope of PQ is 2
∴ Equation of the chord PQ
y – y₁ = m(x – x₁)
y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
On solving the curve and line we get the point Q(4, 5)

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x.
Solution:
To find points of intersection of x² + y² = 16 and y² = 6x are (2, \(2 \sqrt{3})\)) and (2, –\(2 \sqrt{3})\))
Due to symmetrical property,

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8 Additional Problems

Question 1.
Find the area of the region enclosed by y² = x and y = x – 2.
Solution:
The points of intersection of the parabola y² = x and the line y = x – 2 are (1, -1) and (4, 2)

To compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect toy no splitting is necessary.

Question 2.
Find the area bounded by the curve y = x³ and the line y = x.
Solution:
The line y = x lies above the curve y = x³ in the first quadrant and y = x³ lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x³, y = x ⇒ x³ = x. We get x = {0, ± 1}

Question 3.
Find the area of the loop of the curve 3ay² = x (x – a)².
Solution:
Put y = 0; we get x = 0, a
It meets the x – axis at x = 0 and x = a
∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 4.
Find the area between the line y = x + 1 and the curve y = x² – 1.
Solution:
To get the points of intersection of the curves we should solve the equations y = x +1 and y = x² – 1.
we get, x² – 1 = x + 1
x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0
x = – 1 or x = 2
∴ The line intersects the curve at x = – 1 and x = 2.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3

Question 1.

Solution:

Question 2.
For any vector \(\vec{a}\), prove that
Solution:

Question 3.
Prove that \([\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}]\) = 0
Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.
If \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) are coplanar vectors, show that \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}\)
Solution:

Question 7.
If , find the values of l, m, n
Solution:

On solving (3) & (4)

Question 8.
If \(\hat{a}, \hat{b}, \hat{c}\) are three unit vectors such that \(\hat{b} \text { and } \hat{c}\) are non-parallel and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\), find the angle between \(\hat{a}\) and \(\hat{c}\).
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3 Additional Problems

Question 1.
If and show that they are not equal.
Solution:

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2

Question 1.

Solution:

Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors .
Solution:
Volume of the parallelepiped = \(\| \vec{a}, \vec{b}, \vec{c}]\)

= -264 + 224 + 760 = 720 cubic units

Question 3.
The volume of the parallelepiped whose coterminus edges are , \(-3 \vec{i}+7 \vec{j}+5 \vec{k}\) is 90 cubic units. Find the value of λ
Solution:
Given, Volume of the parallelepiped = 90 cubic units

Question 4.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) be the concurrent edges of parallelepiped
Given volume of parallelepiped = 4 cubic units

Question 5.
Find the altitude of a parallelepiped determined by the vectors \(\vec{a}=-2 \hat{i}+5 \hat{j}+3 \hat{k}\), \(\hat{b}=\hat{i}+3 \hat{j}-2 \hat{k}\) and \(\vec{c}=-3 \vec{i}+\vec{j}+4 \vec{k}\) if the base is taken as the parallelogram determined by b and c.
Solution:
Volume = Base Area × Height
\(|[\vec{a}, \vec{b}, \vec{c}]|=|\vec{b} \times \vec{c}|\) × Height

Question 6.
Determine whether the three vectors \(2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+2 \hat{k}\) and \(3 \hat{i}+\hat{j}+3 \hat{k}\) are coplanar.
Solution:

Question 7.
Let If c₁ = 1 and c₂ = 2, find c₃ such that \(\vec{a}, \vec{b}\) and \(\vec{c}\) and c are coplanar.
Solution:

Question 8.
If , show that \([\vec{a} \vec{b} \vec{c}]\) depends neither x nor y.
Solution:

Question 9.
If the vectors

are coplanar, prove that c is the geometric mean of a and b.
Solution:

∴ c is the geometric means of ‘a’ and ‘b’.

Question 10.
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-zero vectors such that \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\) show that .
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2 Additional Problems

Question 1.
If the edges meet a vertex, find the volume of the parallelepiped.
Solution:
Volume of the parallelepiped =
The volume cannot be negative
∴ Volume of parallelepiped = 264 cu. units

Question 2.
If and \(\vec{x} \neq \overrightarrow{0}\) then show that \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.
Solution:

Question 3.
The volume of a parallelepiped whose edges are represented by \(-12 \vec{i}+\lambda k\), \(3 \vec{j}-\vec{k}, 2 \vec{i}+\vec{j}-15 \vec{k}\) is 546. Find the value of λ.
Solution:
Volume of the parallelepiped =
= -12 [-45 + 1] – 0 () + λ [0 – 6] = -12 (-44) -6 λ
= 528 – 6λ = 546 (given)
⇒ -6λ = 546 – 528 = 18
∴ λ = \(\frac{18}{-6}\) = -3

Question 4.
Prove that \(|\vec{a} \vec{b} \vec{c}|\) = abc if and only if \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular.
Solution:

Question 5.
Show that the points (1, 3, 1), (1, 1, -1), (-1, 1, 1), (2, 2, -1) are lying on the same plane.
Solution:

Question 6.

Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following

Solution:
We know that

(ii)
Solution:

= 0 + 1 + 24 + 4 = 29

Question 2.

Solution:

By using Gamma integral (n = 1; a = α)

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6

Question 1.

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:
We know that,

(viii)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6 Additional Problems

Question 1.
Evaluate

Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

Question 2.

Solution: