MHT CET · Maths · Vector Algebra
If the vectors \(\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}\) and \(\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}\) are collinear, then \(\mathrm{p}\) and \(\mathrm{q}\) are
- A \(\mathrm{p}=3, \mathrm{q}=-2\)
- B \(\mathrm{p}=3, \mathrm{q}=\mathrm{n} 12\)
- C \(\mathrm{p}=-3, \mathrm{q}=12\)
- D \(\mathrm{p}=-3, \mathrm{q}=-12\)
Answer & Solution
Correct Answer
(C) \(\mathrm{p}=-3, \mathrm{q}=12\)
Step-by-step Solution
Detailed explanation
Let \(\vec{a}=x \vec{b}\)
\(\therefore 2 \hat{i}+p \hat{j}+4 \hat{k}=6 x \hat{i}-9 x \hat{j}+q x \hat{j} \)
\( \therefore 2=6 x \Rightarrow x=\frac{1}{3} \)
\( p=-9 x \Rightarrow(-9)\left(\frac{1}{3}\right)=-3 \text { and } 4=q x=q\) \(\left(\frac{1}{3}\right) \Rightarrow q=4(3)=12\)
\(\therefore 2 \hat{i}+p \hat{j}+4 \hat{k}=6 x \hat{i}-9 x \hat{j}+q x \hat{j} \)
\( \therefore 2=6 x \Rightarrow x=\frac{1}{3} \)
\( p=-9 x \Rightarrow(-9)\left(\frac{1}{3}\right)=-3 \text { and } 4=q x=q\) \(\left(\frac{1}{3}\right) \Rightarrow q=4(3)=12\)
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