Prove the following

Question:

Let $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ has the magnitude 12 then one such vector is

  1. $4(2 \hat{i}+2 \hat{j}-\hat{k})$

  2. $4(-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

  3. $4(2 \hat{i}-2 \hat{j}-\hat{k})$

  4. $4(2 \hat{i}+2 \hat{j}+\hat{k})$


Correct Option: , 3

Solution:

$(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})$

$=2(\vec{b} \times \vec{a})$

$=2\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -2 \\ 3 & 2 & 2\end{array}\right|$

$=2(8 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$Required vector $=\pm 12 \frac{(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{3}$

$=\pm 4(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$

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