RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.3

RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.3 are part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Solutions for RBSE Solutions for Class 12 Maths Chapter 13 Vector Ex 13.3.

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.3

Question 1.
Find vector product of vectors

and

Solution:

Question 2.
Find perpendicular unit vector of vectors

and

Solution:

Question 3.
For vectors \(\overrightarrow { a } \) and \(\overrightarrow { b } \), prove that

Solution:

Question 4.
Prove that

Solution:
According to question,

Question 5.
If \(\overrightarrow { a } \), \(\overrightarrow { b } \), \(\overrightarrow { c } \) are unit vectors, such that
\(\overrightarrow { a } \) . \(\overrightarrow { b } \) = 0 = \(\overrightarrow { a } \) . \(\overrightarrow { c } \) and angle between \(\overrightarrow { b } \) and \(\overrightarrow { c } \) is \(\frac { \pi }{ 6 } \), then prove that \(\overrightarrow { a } \) = ± 2 (\(\overrightarrow { b } \) × \(\overrightarrow { c } \))
Solution:
Given that

Question 6.
Find the value of

Solution:
We know that if \(\overrightarrow { a } \) and \(\overrightarrow { b } \) are two vectors and θ is the angle between them, then

Question 7.
Find vector perpendicular to the vectors

and

whose magnitude is 9 unit.
Solution:

Question 8.
Show that:

also, explain geometrically.
Solution:

= 2(vector area of parallelogram ABCD).
Thus we conclude that area of parallelogram whose adjacent sides are diagonals of given parallelogram is twice the area of given parallellogram.