MHT CET · Maths · Vector Algebra
The vector projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\), where \(A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)\) and \(\mathrm{D} \equiv(7,0,10)\), is
- A \(\frac{1}{49}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)
- B \(\frac{1}{6}(-\hat{i}-\hat{j}-2 \hat{k})\)
- C \(-\frac{1}{49}(3 \hat{i}-6 \hat{j}+2 \hat{k})\)
- D \(-\frac{1}{6}(-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{49}(3 \hat{i}-6 \hat{j}+2 \hat{k})\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\
& \overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
Vector projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\)
\(\begin{aligned}
& =(\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}) \frac{\overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}|^2} \\
& =(-3+6-4) \frac{(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\left(\sqrt{3^2+(-6)^2+2^2}\right)^2} \\
& =\frac{-1}{49}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{aligned}\)
& \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\
& \overline{\mathrm{CD}}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
Vector projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\)
\(\begin{aligned}
& =(\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}) \frac{\overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}|^2} \\
& =(-3+6-4) \frac{(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\left(\sqrt{3^2+(-6)^2+2^2}\right)^2} \\
& =\frac{-1}{49}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{aligned}\)
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