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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
Question
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 $.....$
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Answer

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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