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

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-10 (Vector Algebra)Miscellaneous Exercise 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.

Q1. Write down a unit vector in KY-plane, making an angle of ${30}^{\circ }$ with the positive direction of x-axis.

Q2.

Answer.  $\begin{array}{rl}& =\left({x}_{2}-{x}_{1}\right)\stackrel{^}{i}+\left({y}_{2}-{y}_{1}\right)\stackrel{^}{j}+\left({z}_{2}-{z}_{1}\right)\stackrel{^}{k}\\ |\stackrel{\to }{\mathrm{P}\mathrm{Q}}|& =\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}\end{array}$

Q3. A girl walks 4 km towards west, then she walks 3 km in a direction ${30}^{\circ }$east of north and stops. Determine the girl's displacement from her initial point of departure.

Answer. Let O and B be the initial and final positions of the girl respectively. Then, the girl's position can be shown as:  $\begin{array}{l}\stackrel{\to }{\mathrm{O}\mathrm{A}}=-4\stackrel{^}{i}\\ \stackrel{\to }{\mathrm{A}\mathrm{B}}=\stackrel{^}{i}|\stackrel{\to }{\mathrm{A}\mathrm{B}}|\mathrm{cos}{60}^{\circ }+\stackrel{^}{j}|\stackrel{\to }{\mathrm{A}\mathrm{B}}|\mathrm{sin}{60}^{\circ }\end{array}$ $\begin{array}{l}=\stackrel{^}{i}3×\frac{1}{2}+\stackrel{^}{j}3×\frac{\sqrt{3}}{2}\\ =\frac{3}{2}\stackrel{^}{i}+\frac{3\sqrt{3}}{2}\stackrel{^}{j}\end{array}$ $\begin{array}{rl}\stackrel{\to }{\mathrm{O}\mathrm{B}}& =\stackrel{\to }{\mathrm{O}\mathrm{A}}+\stackrel{\to }{\mathrm{A}\mathrm{B}}\\ & =\left(-4\stackrel{^}{i}\right)+\left(\frac{3}{2}\stackrel{^}{i}+\frac{3\sqrt{3}}{2}\stackrel{^}{j}\right)\\ & =\left(-4+\frac{3}{2}\right)\stackrel{^}{i}+\frac{3\sqrt{3}}{2}\stackrel{^}{j}\\ & =\left(\frac{-8+3}{2}\right)\stackrel{^}{i}+\frac{3\sqrt{3}}{2}\stackrel{^}{j}\\ & =\frac{-5}{2}\stackrel{^}{i}+\frac{3\sqrt{3}}{2}\stackrel{^}{3}\end{array}$

Q4. If

Answer. (as shown in the following figure)  Now, by the triangle law of vector addition we have $\stackrel{\to }{a}=\stackrel{\to }{b}+\stackrel{\to }{c}$ It is clearly known that Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side. $\therefore |\overline{a}|<|\stackrel{\to }{b}|+|\overline{c}|$ Hence it is not true that

Q5. Find the value of x  is a unit vector .

Answer.  $\begin{array}{l}\stackrel{\to }{c}=\stackrel{\to }{a}+\stackrel{\to }{b}=\left(2+1\right)\stackrel{^}{i}+\left(3-2\right)\stackrel{^}{j}+\left(-1+1\right)\stackrel{^}{k}=3\stackrel{^}{i}+\stackrel{^}{j}\\ \therefore |\stackrel{\to }{c}|=\sqrt{{3}^{2}+{1}^{2}}=\sqrt{9+1}=\sqrt{10}\\ \therefore \stackrel{^}{c}=\frac{\stackrel{\to }{c}}{|\stackrel{\to }{c}|}=\frac{\left(3\stackrel{^}{i}+\stackrel{^}{j}\right)}{\sqrt{10}}\end{array}$
Answer.  $\begin{array}{rl}& =2\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}-2\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}+3\stackrel{^}{i}-6\stackrel{^}{j}+3\stackrel{^}{k}\\ & =3\stackrel{^}{i}-3\stackrel{^}{j}+2\stackrel{^}{k}\\ |2\stackrel{\to }{a}-\stackrel{\to }{b}+3c|=\sqrt{{3}^{2}+\left(-3{\right)}^{2}+{2}^{2}}& =\sqrt{9+9+4}=\sqrt{22}\end{array}$