# NCERT Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-10 (Vector Algebra)Exercise 10.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 10.3

Q1.

Answer.  Hence , the angle between the given vectors

Q2. Find the angle between the given vector Q3. Find the projection of the vector

Answer.  Hence the projection of vector

Q4. Find the projection of the vector

Q5. Show that each of the given three vectors is a unit vector: $\frac{1}{7}\left(2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}\right)\cdot \frac{1}{7}\left(3\stackrel{^}{i}-6\stackrel{^}{j}+2\stackrel{^}{k}\right),\frac{1}{7}\left(6\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}\right)$. Also, show that they are mutually perpendicular to each other.

Q6. Find

Answer. $\begin{array}{l}\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)\cdot \left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)=8\\ ⇒\stackrel{\to }{a}\cdot \stackrel{\to }{a}-\stackrel{\to }{a}\stackrel{\to }{b}+\stackrel{\to }{b}\cdot \stackrel{\to }{a}-\stackrel{\to }{b}\cdot \stackrel{\to }{b}=8\\ ⇒|\stackrel{\to }{a}{|}^{2}-|\stackrel{\to }{b}{|}^{2}=8\\ ⇒\left(8|\stackrel{\to }{b}|{\right)}^{2}-|\stackrel{\to }{b}{|}^{2}=8\\ ⇒64|\stackrel{\to }{b}{|}^{2}-|\stackrel{\to }{b}{|}^{2}=8\\ ⇒63|\stackrel{\to }{b}{|}^{2}=8\end{array}$

Q7. Evaluate the product

Q8. Find the magnitude of two vectors  , having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is $\frac{1}{2}$.