3D Geometry 3 Question 13
####13. The magnitude of the projection of the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ on the vector perpendicular to the plane containing the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, is
(2019 Main, 8 April I)
(a) $\frac{\sqrt{3}}{2}$
(b) $\sqrt{6}$
(c) $3 \sqrt{6}$
(d) $\sqrt{\frac{3}{2}}$
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Answer:
(d)
Solution:
- The normal vector to the plane containing the vectors $(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ is $\mathbf{n}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$
$ \begin{aligned} & \begin{array}{lll} \hat{\mathbf{i}} \quad \hat{\mathbf{j}} \quad \hat{\mathbf{k}} \end{array} \\ = & 1 \quad 1 \quad 1 \\ & 1 \quad 2 \quad 3 \\ & =\hat{\mathbf{i}}(3-2)-\hat{\mathbf{j}}(3-1)+\hat{\mathbf{k}}(2-1)=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $
Now, magnitude of the projection of vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ on normal vector $\mathbf{n}$ is
$ \begin{aligned} \frac{|(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot \mathbf{n}|}{|\mathbf{n}|} & =\frac{|(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-2 \hat{j}+\hat{\mathbf{k}})|}{\sqrt{1+4+1}} \\ & =\frac{|2-6+1|}{\sqrt{6}}=\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}} \text { units } \end{aligned} $
