If $A=(1,8,4), B=(2,-3,1)$, then the direction cosines of a normal to the plane $A O B$ is

If $A=(1,8,4), B=(2,-3,1)$, then the direction cosines of a normal to the plane $A O B$ is
  1. $\frac{2}{\sqrt{78}}, \frac{5}{\sqrt{78}}, \frac{-7}{\sqrt{78}}$
  2. $\frac{2 \sqrt{10}}{9}, \frac{7 \sqrt{10}}{90}, \frac{-19 \sqrt{10}}{90}$
  3. $\frac{4}{\sqrt{218}}, \frac{9}{\sqrt{218}}, \frac{-11}{\sqrt{218}}$
  4. $\frac{2}{11}, \frac{6}{11}, \frac{-9}{11}$

Solution

Given, $A=(1,8,4)$ and $B=(2,-3,1)$ $ \begin{aligned} & \therefore \quad \mathbf{O A}=\hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & \text { and } \quad \mathbf{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \text { Now, } \quad \hat{\mathbf{n}}=\frac{\mathbf{O A} \times \mathbf{O B}}{|\mathbf{O A} \times \mathbf{O B}|} \\ & \end{aligned} $ Here, $\mathbf{O A} \times \mathbf{O B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 8 & 4 \\ 2 & -3 & 1\end{array}\right|$ $ \begin{aligned} & =\hat{\mathbf{i}}(8+12)-\hat{\mathbf{j}}(1-8)+\hat{\mathbf{k}}(-3-16)=20 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-19 \hat{\mathbf{k}} \\ & \text { and }|\mathbf{O A} \times \mathbf{O B}|=\sqrt{(20)^2+7^2+(-19)^2} \\ & =\sqrt{400+49+361}=\sqrt{810}=9 \sqrt{10} \\ & \therefore \quad \hat{\mathbf{n}}=\frac{\mathbf{2 0} \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-19 \hat{\mathbf{k}}}{9 \sqrt{10}} \\ & \end{aligned} $ $\therefore$ Direction cosines are $\frac{20}{9 \sqrt{10}}, \frac{7}{9 \sqrt{10}}, \frac{-19}{9 \sqrt{10}}$ i.e. $\quad \frac{2 \sqrt{10}}{9}, \frac{7 \sqrt{10}}{90}, \frac{-19 \sqrt{10}}{90}$

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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