Question
The unit vector perpendicular to the plane passing through points $\text{P}\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big),\text{Q}\big(2\hat{\text{i}}-\hat{\text{k}}\big)$ and $\text{R}\big(2\hat{\text{j}}+\hat{\text{k}}\big)$ is:

  1. $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$

  2. $\sqrt{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

  3. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

  4. $\frac{1}{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

Answer

  1. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

Solution:

The vector $\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$ is perpendicular to the vectors $\overrightarrow{\text{PQ}}$  and $\overrightarrow{\text{PR}}.$

$\therefore$ Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$

Now,

$\overrightarrow{\text{PQ}}=\text{P.V}\text{ of }\text{Q}-\text{P.V}.\text{ of P}$

$=\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$

$\overrightarrow{\text{PR}}=\text{P.V}\text{ of }\text{R}-\text{P.V}.\text{ of P}$

$=-\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$

$\therefore\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&-3\\-1&3&-1 \end{vmatrix}$

$=8\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$

$=4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

$\Rightarrow\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|=\sqrt{64+16+16}$

$=\sqrt{96}$

$=4\sqrt{6}$

Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$

$=\frac{4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{4\sqrt{6}}$

$=\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

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