Q144 / 155 · WBJEE 2017 · Physics
Three vectors $\mathbf{A}=a \mathbf{i}+\mathbf{j}+\mathbf{k} ; \mathbf{B}=\mathbf{i}+b \mathbf{j}+\mathbf{k}$
and $\mathbf{C}=\mathbf{i}+\mathbf{j}+c \mathbf{k}$ are mutually perpendicular (i, $\mathbf{j}$ and $\mathbf{k}$ are unit vectors along $X, Y$ and $Z$ axes respectively). The respective values of $a$, $b$ and $c$ are
- A0,0,0
- Correct answerB$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$
- C1,-1,1
- D$\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$
Solution
Three vectors are mutually perpendicular, so the scalar or dot product of two vectors will zero. So,
$\mathbf{A} \cdot \mathbf{B}=0, \mathbf{B} \mathbf{C}=0,$
So, $\mathbf{A} \cdot \mathbf{B}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}})=0$
$\Rightarrow a+b+1=0$
$\mathbf{B} \cdot \mathbf{C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}=$
$\Rightarrow 1+b+c=0$
$\mathbf{A} \cdot \mathbf{C}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}})=0$
$\mathbf{A} \cdot \mathbf{C}=0$
$\Rightarrow \quad a+1+c=0$
(iii Addition of Eqs. (i), (ii) and (iii),
$\begin{array}{l}
2(a+b+c)+3=0 \\
a+b+c=-\frac{3}{2}
\end{array}$
$\Rightarrow \quad-1+c=-\frac{3}{2}$
$\begin{aligned}
c &=-\frac{3}{2}+1 \\
&=-\frac{1}{2}=0.5 \\
a &=-\frac{1}{2} \\
b &=-\frac{1}{2} \\
c &=-\frac{1}{2}
\end{aligned}$
Hence, the correct answer is option: 'B'.