Let $\mathbf{m}$ be the unit vector orthogonal to the vector…

1. $\frac{1}{\sqrt{26}}$
2. $\frac{1}{\sqrt{5}}$
3. $\frac{5}{\sqrt{26}}$
4. 1

Solution

The vector $\mathbf{m}$ is coplane with $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(\hat{\mathbf{j}}-\hat{\mathbf{k}})$ So, $ \begin{aligned} & \mathbf{m}=x(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})+y(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & \mathbf{m}=2 x \hat{\mathbf{i}}+(x+y) \hat{\mathbf{j}}-y \hat{\mathbf{k}} \end{aligned} $ $\therefore \mathbf{m}$ is orthogonal to the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, so $ 2 x-(x+y)-y=0 $
$ \begin{array}{lrl} \therefore & |\mathbf{m}|=1 \Rightarrow 4 x^2+(x+y)^2+y^2=1 \\ \Rightarrow & 16 y^2+9 y^2+y^2=1 \\ \Rightarrow & 26 y^2=1 \Rightarrow y= \pm \frac{1}{\sqrt{26}} . \end{array} $ So, $ x= \pm \frac{2}{\sqrt{26}} $ $ \therefore \quad \mathbf{m}= \pm\left(\frac{4}{\sqrt{26}} \hat{\mathbf{i}}+\frac{3}{\sqrt{26}} \hat{\mathbf{j}}-\frac{1}{\sqrt{26}} \hat{\mathbf{k}}\right) $ The length of the perpendicular from the origin to the plane $\mathbf{r} . \mathbf{m}=\mathbf{a} \cdot \mathbf{m}$ is So, $ \begin{aligned} & \mathbf{m}|=| \hat{\mathbf{i}}-\hat{\mathbf{k}}) \cdot\left[\frac{4}{\sqrt{26}} \hat{\mathbf{i}}+\frac{3}{\sqrt{26}} \hat{\mathbf{j}}-\frac{1}{\sqrt{26}} \hat{\mathbf{k}}\right] \mid \\ & =\frac{4}{\sqrt{26}}+\frac{1}{\sqrt{26}}=\frac{5}{\sqrt{26}} \text { units. } \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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