Let p=2 i+3 j+k and q=i+2 j+k be two vectors. If a vector r=(a i+… - Vidyadip

MCQ

Let $\vec{p}=2 \hat{i}+3 \hat{j}+k$ and $\vec{q}=\hat{i}+2 \hat{j}+k$ be two vectors. If $a$ vector $\vec{r}=(a \hat{i}+\beta \hat{j}+\gamma k)$ is perpendicular to each of the vectors $(\vec{p}+\bar{q})$ and $(\vec{p}-\vec{q})$, and $|\vec{r}|=\sqrt{3}$, then $|\alpha|+|\beta|+|\gamma|$ is equal to $.....$

✓
$3$
B
$4$
C
$1$
D
$2$

✓

Answer

Correct option: A.

$3$

a
$\overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { (Given) }$

$\overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

Now $(\vec{p}+\vec{q}) \times(\vec{p}-\vec{q})=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0\end{array}\right|$

$=2 \hat{i}-2 \hat{j}-2 \hat{k}$

$\Rightarrow \vec{r}=\pm \sqrt{3} \frac{((\vec{p}+\vec{q}) \times(\vec{p}-\vec{q}))}{|(\vec{p}+\vec{q}) \times(\vec{p}-\vec{q})|}=\pm \frac{\sqrt{3}(-2 \hat{i}-2 \hat{j}-2 \hat{k})}{\sqrt{2^{2}+2^{2}+2^{2}}}$

$\vec{r}=\pm(-\hat{i}-\hat{j}-\hat{k})$

According to question

$\vec{r}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

$\text { So }|\alpha|=1,|\beta|=1,|\gamma|=1$

$\Rightarrow|\alpha|+|\beta|+|\gamma|=3$

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