MHT CET · Maths · Vector Algebra
The vector projection of \(\overline{P Q}\) on \(\overline{A B}\), where \(P \equiv(-2,1,3), Q \equiv(3,2,5), A \equiv(4,-3,5)\) and \(B \equiv(7,-5,-1)\) is
- A \(\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})\)
- B \(\frac{1}{7}(3 \hat{i}-2 \hat{j}+6 \hat{k})\)
- C \((3 \hat{i}-2 \hat{j}-6 \hat{k})\)
- D \(\frac{1}{7}(3 \hat{i}-2 \hat{j}-6 \hat{k})\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \overline{P Q}=(3+2) \hat{i}+(2-1) \hat{j}+(5-3) \hat{k}=5 \hat{i}+\hat{j}+2 \widehat{k} \\
& \overline{A B}=(7-4) \hat{i}+(-5+3) \hat{j}+(-1-5) \hat{k}=3 \hat{i}-2 \hat{j}-6 \hat{k}
\end{aligned}\)
now vector projection of \(\overrightarrow{P Q}\) on \(\overrightarrow{A B}\) is \(\frac{(\overrightarrow{P Q} \cdot \overrightarrow{A B}) \overrightarrow{A B}}{|\overrightarrow{A B}|^2}\)
\(=\frac{\{5 \times 3+1 \times(-1)+2 \times(-6)\}(3 \hat{i}-2 \hat{j}-6 \hat{k})}{3^2+(-2)^2+(-6)^2}=\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})\)
& \overline{P Q}=(3+2) \hat{i}+(2-1) \hat{j}+(5-3) \hat{k}=5 \hat{i}+\hat{j}+2 \widehat{k} \\
& \overline{A B}=(7-4) \hat{i}+(-5+3) \hat{j}+(-1-5) \hat{k}=3 \hat{i}-2 \hat{j}-6 \hat{k}
\end{aligned}\)
now vector projection of \(\overrightarrow{P Q}\) on \(\overrightarrow{A B}\) is \(\frac{(\overrightarrow{P Q} \cdot \overrightarrow{A B}) \overrightarrow{A B}}{|\overrightarrow{A B}|^2}\)
\(=\frac{\{5 \times 3+1 \times(-1)+2 \times(-6)\}(3 \hat{i}-2 \hat{j}-6 \hat{k})}{3^2+(-2)^2+(-6)^2}=\frac{1}{49}(3 \hat{i}-2 \hat{j}-6 \hat{k})\)
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