RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.2

RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.2 are part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.12.

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.2

Question 1.
If magnitude of two vectors be 4 and 5 units, then find scalar product of them, where angle between them be :
(i) 60°
(ii) 90°
(iii) 30°
Solution:

Question 2.
Find \(\overrightarrow { a } \). \(\overrightarrow { b } \) , if \(\overrightarrow { a } \) and \(\overrightarrow { b } \) are as follow :


Solution:

Question 3.
Prove that:

Solution:

Question 4.
If coordinates of P and Q are (3, 4) and (12, 4) respectively, then find ∠POQ where O is origin.
Solution:

Question 5.
For which value of λ, vectors \(\overrightarrow { a } \) and \(\overrightarrow { b } \) are mutually perpendicular :

Solution:

Question 6.
Find the projection of the vector

on the vector

Solution:

Question 7.
If

and

then find a vector \(\overrightarrow { c } \), so that \(\overrightarrow { a } \), \(\overrightarrow { b } \), \(\overrightarrow { c } \) represents the sides of a right angled triangle.
Solution:
Given that

Question 8.
If | \(\overrightarrow { a } \) + \(\overrightarrow { b } \) | = | \(\overrightarrow { a } \) – \(\overrightarrow { b } \) |, then prove that \(\overrightarrow { a } \) and \(\overrightarrow { b } \) are mutually perpendicular vectors.
Solution:
According to question,

Question 9.
If coordinates of points A, B, C and D are (3, 2, 4), (4, 5, -1), (6, 3, 2) and (2, 1, 0) respectively, then prove that lines \(\overrightarrow { AB } \) and \(\overrightarrow { CD } \) are mutually erpendicular.
Solution:
Given that coordinates of points A, B, C and D are (3,2,4), (4,5,-1), (6,3,2) and (2,1,0) respectively.
Then position vectors of A, B, C and D with respect to origin are

Question 10.
For any vector \(\overrightarrow { a } \), prove that

Solution:

Question 11.
Use vectors to prove that sum of square of diagonal of a parallelogram is equal to the sum of square of their side.
Solution:
Let OACB is a parallelogram. Taking O as origin, the position vectors of A and B are \(\overrightarrow { a } \) and \(\overrightarrow { b } \) respectively.

∴ Sum of the squares of diagonals = sum of the squares of sides.
Hence the sum of square of diagonals of a parallelogram is equal to the sum of square of their sides.