If $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$ are orthogonal and $|\vec{a}|=|\vec{b}|$,then $(\lambda, \mu) = $

  • A
    $\left(\frac{1}{4}, \frac{7}{4}\right)$
  • B
    $\left(\frac{7}{4}, \frac{1}{4}\right)$
  • C
    $\left(\frac{1}{4}, \frac{9}{4}\right)$
  • D
    $\left(-\frac{1}{4}, \frac{9}{4}\right)$

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Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{a}$ is equal to ...........

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then $\cos ^2 A$ is equal to

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