Q

# Can someone explain - Vector Algebra - JEE Main-2

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 magnitude $12$ then one such vector is :

• Option 1)

$4\left ( -2\hat{i}-2\hat{j}+\hat{k} \right )$

• Option 2)

$4\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )$

• Option 3)

$4\left ( 2\hat{i}+2\hat{j}-\hat{k} \right )$

• Option 4)

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

Views

$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ ,                      $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$

$\left ( \vec{a} +\vec{b}\right )\times \left ( \vec{a}-\vec{b} \right )=2\left ( \vec{b} \times \vec{a}\right )$

$=2\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &-2 \\ 3 &2 &2 \end{vmatrix}$

$=2\left ( 8\hat{i}-8\hat{j}+4\hat{k} \right )$

$=8\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )$

$\hat{h}=\frac{\vec{\alpha }\times \vec{\beta }}{\left | \vec{\alpha }\times \vec{\beta } \right |}=\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}$

required vector = $\pm 12\cdot \frac{\left ( 2\hat{i}-2\hat{j}-\hat{j} \right )}{3}$

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

Option 1)

$4\left ( -2\hat{i}-2\hat{j}+\hat{k} \right )$

Option 2)

$4\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )$

Option 3)

$4\left ( 2\hat{i}+2\hat{j}-\hat{k} \right )$

Option 4)

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

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