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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:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 1

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:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 2

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:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 3

(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).

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

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
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 4
Note: Selection of \(\vec{a}\) and \(\vec{b}\) is your choice.

Question 5.
Find the acute angle between the following lines.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 5
(iii) 2x = 3y = -z and 6x = -y = -4z
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 6
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 7

Question 6.
The vertices of ∆ABC are A(7, 2, 1), B(6, 0, 3), and C(4, 2, 4) . Find ∠ABC.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 8

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:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 9
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 10

Question 8.
If the straight lines Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 11 are perpendicular to each other, find the value of m.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 12

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

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
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 13

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
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 14

Question 3.
Find the angle between the lines.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 15
Solution:
Let the given lines be in the direction of \(\vec{u}\) and \(\vec{v}\)
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 16

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

Question 4.
Find the angle between the following lines Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 17
Solution:
Angle between two lines is the same as angle between their parallel vectors.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 18

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