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Question: Answered & Verified by Expert
If $P \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}$ are the position vectors of three collinear points $A, B$ and $C$ respectively, then the vector in the direction of $\mathbf{A B}$ of length $|P|$ units is
MathematicsVector AlgebraTS EAMCETTS EAMCET 2021 (06 Aug Shift 2)
Options:
  • A $\frac{2}{5 \sqrt{3}}(\hat{i}+5 \hat{j}-7 \hat{k})$
  • B $\frac{1}{\sqrt{83}}(3 \hat{i}+5 \hat{j}-7 \hat{k})$
  • C $\frac{1}{\sqrt{78}}(2 \hat{i}+5 \hat{j}-7 \hat{k})$
  • D $\frac{1}{5 \sqrt{3}}(\hat{i}+5 \hat{j}-7 \hat{k})$
Solution:
2260 Upvotes Verified Answer
The correct answer is: $\frac{1}{5 \sqrt{3}}(\hat{i}+5 \hat{j}-7 \hat{k})$
Let $\mathbf{a}=P \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$,
$\mathbf{b}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $\mathbf{c}=4 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}$
Since, $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are collinear.
$\begin{aligned}
& \therefore \quad[\mathbf{a} \mathbf{b} \mathbf{c}]=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
P & -2 & 3 \\
2 & 3 & -4 \\
4 & 13 & -18
\end{array}\right|=0 \\
& \Rightarrow \quad-2 P-40+42=0 \\
& \Rightarrow \quad P=1 \\
& \therefore \quad \mathbf{A B}=\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\
& \text { and } \\
& |\mathbf{A B}|=\sqrt{75}=5 \sqrt{3} \\
&
\end{aligned}$
Then, the vector on the direction of
$\mathbf{A B}=\frac{1}{5 \sqrt{3}}(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})$

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