Let $\vec{a} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}$. The projection of the sum of the vectors $\vec{a}$ and $\vec{b}$ on the vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is:

  • A
    $0$
  • B
    $4 \sqrt{2}$
  • C
    $7 \sqrt{2}$
  • D
    $\frac{1}{\sqrt{2}}$

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Similar Questions

The acute angle $\theta$ between the vector $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and the plane containing the vectors $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ is:

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is

The area of the parallelogram for which the vectors $\hat{i}+\hat{j}+2 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ are adjacent sides is equal to

If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them,then for what value of $\theta$ is $2\hat{u} \times 3\hat{v}$ a unit vector?

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.

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