If $\mathrm{A}=(-2,2,3), \mathrm{B}=(3,2,2), \mathrm{C}=(4,-3,5)$ and $\mathrm{D}=(7,-5,-1)$ Then the…

If $\mathrm{A}=(-2,2,3), \mathrm{B}=(3,2,2), \mathrm{C}=(4,-3,5)$ and $\mathrm{D}=(7,-5,-1)$ Then the projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{CD}}$ is
  1. 4
  2. 3
  3. $\frac{12}{\sqrt{7}}$
  4. None of these

Solution

$\begin{aligned} & \mathrm{A}=(-2,2,3) ; \mathrm{B}=(3,2,2) ; \mathrm{C}=(4,-3,5) \text { and } \mathrm{D}=(7,-5,-1) \\ & \overline{\mathrm{AB}}=5 \hat{\mathrm{i}}-\hat{\mathrm{k}} \text { and } \overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & \text { Projection of } \overline{\mathrm{AB}} \text { on } \overline{\mathrm{CD}} \\ & =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}}{|\mathrm{CD}|}=\frac{(5 \hat{\mathrm{i}}-\hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})}{\sqrt{(3)^2+(-2)^2+(-6)^2}}=\frac{15+6}{7}=3\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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