Search any question & find its solution
Question:
Answered & Verified by Expert
The point collinear with $(1,-2,-3)$ and $(2,0,0)$ among the following is
Options:
Solution:
1796 Upvotes
Verified Answer
The correct answer is:
$(0,-4,-6)$
Let $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}$
Now take option (c).
Let $\mathbf{c}=0 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$
Now, a
$\begin{aligned}
& \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}
1 & -2 & -3 \\
2 & 0 & 0 \\
0 & -4 & -6
\end{array}\right| \\
& =1(0)+2(-12)-3(-8)=-24+24=0
\end{aligned}$
$\because$ Option (c) is correct.
Now take option (c).
Let $\mathbf{c}=0 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$
Now, a
$\begin{aligned}
& \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}
1 & -2 & -3 \\
2 & 0 & 0 \\
0 & -4 & -6
\end{array}\right| \\
& =1(0)+2(-12)-3(-8)=-24+24=0
\end{aligned}$
$\because$ Option (c) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.