Q. Given, $\overrightarrow{\mathbf{p}} = 3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}},\ \overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}},\ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}},\ \overrightarrow{\mathbf{c}} = \widehat{\dot{\mathbf{i}}} + \widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{p}} = x\overrightarrow{\mathbf{a}} + \ y\overrightarrow{\mathbf{b}} + z\overrightarrow{\mathbf{\ c}}$ , then $x,\ y,\ z$ are respectively

A) $\frac{3}{2}\ ,\frac{1}{2}\ ,\frac{5}{2}$
B) $\frac{1}{2}\ ,\frac{3}{2}\ ,\frac{5}{2}$
C) $\frac{5}{2}\ ,\frac{3}{2}\ ,\frac{1}{2}$
D) $\frac{1}{2}\ ,\frac{5}{2}\ ,\frac{3}{2}$
Explanation

(b)
$\overrightarrow{\mathbf{p}} = x\overrightarrow{\mathbf{a}} + y\overrightarrow{\mathbf{b}} + z\overrightarrow{\mathbf{c}}$
$\Longrightarrow 3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 4\widehat{\dot{\mathbf{k}}} = x\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) + y\left( \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) + z\left( \widehat{\dot{\mathbf{i}}} + \widehat{\mathbf{k}} \right)$
$\Longrightarrow 3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}} = (x + z)\widehat{\dot{\mathbf{i}}} + (x + y)\widehat{\dot{\mathbf{j}}} + (y + z)\widehat{\mathbf{k}}$
On comparing both sides the coefficients of $\widehat{\dot{\mathbf{i}}},\ \widehat{\dot{\mathbf{j}}},\widehat{\mathbf{k}}$, we get
$x + z = 3\ \ldots.(i)$
$x + y = 2\ldots.(ii)$
$and\ y + z = 4\ (iii)$
on solving Eqs. (i), (ii) and (iii), we get
$x = \frac{1}{2},y = \frac{3}{2},z = \frac{5}{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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