Vectors Ques 34
- Let $\mathbf{a}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be two vectors. If a vector perpendicular to both the vectors $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ has the magnitude 12 , then one such vector is
(2019 Main, 12 April II)
(a) $4(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
(b) $4(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
(c) $4(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
(d) $4(-2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
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Answer:
Correct Answer: 34.(b)
Solution:
Formula:
Vector Product Of Two Vectors:
- Given vectors are
$$ \mathbf{a}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} $$
Now, vectors $\mathbf{a}+\mathbf{b}=4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ and $\mathbf{a}-\mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}$
$\therefore$ A vector which is perpendicular to both the vectors $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ is
$$ \begin{aligned} (\mathbf{a}+\mathbf{b}) \times & (\mathbf{a}-\mathbf{b})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array}\right| \\ & =\hat{\mathbf{i}}(16)-\hat{\mathbf{j}}(16)+\hat{\mathbf{k}}(-8)=8(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \end{aligned} $$
Then, the required vector along $(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}-\mathbf{b})$ having magnitude 12 is
$\pm 12 \times \frac{8(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{8 \times \sqrt{4+4+1}}= \pm 4(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
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