MHT CET · Maths · Three Dimensional Geometry
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
- A 4
- B 3
- C \(\frac{12}{\sqrt{7}}\)
- D None of these
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
\(\mathrm{A}=(-2,2,3) ; \mathrm{B}=(3,2,2) ; \mathrm{C}=(4,-3,5) \text { and }\) \(\mathrm{D}=(7,-5,-1) \)
\( \overline{\mathrm{AB}}=5 \hat{\mathrm{i}}-\hat{\mathrm{k}} \text { and } \overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \)
\( \text { Projection of } \overline{\mathrm{AB}} \text { on } \overline{\mathrm{CD}} \)
\( =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}}{|\mathrm{CD}|}=\frac{(5 \hat{\mathrm{i}}-\hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})}{\sqrt{(3)^2+(-2)^2+(-6)^2}}=\frac{15+6}{7}=3\)
\( \overline{\mathrm{AB}}=5 \hat{\mathrm{i}}-\hat{\mathrm{k}} \text { and } \overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \)
\( \text { Projection of } \overline{\mathrm{AB}} \text { on } \overline{\mathrm{CD}} \)
\( =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}}{|\mathrm{CD}|}=\frac{(5 \hat{\mathrm{i}}-\hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})}{\sqrt{(3)^2+(-2)^2+(-6)^2}}=\frac{15+6}{7}=3\)
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