$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

  • A
    $3$
  • B
    $2$
  • C
    $1$
  • D
    $4$

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Similar Questions

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

If $a = 2i + k$,$b = i + j + k$ and $c = 4i - 3j + 7k$. If $d \times b = c \times b$ and $d \cdot a = 0$,then $d$ is equal to:

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:

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=-\hat{i}+2 \hat{j}+\hat{k}$,$c=\hat{i}+2 \hat{j}-2 \hat{k}$,$n$ is perpendicular to both $a$ and $b$,and $\theta$ is the angle between $c$ and $n$,then $\sin \theta=$

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