Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

  • A
    $\lambda a$
  • B
    $\lambda b$
  • C
    $\lambda c$
  • D
    $0$

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If $l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines,prove that the line whose direction cosines are proportional to $l_{1}+l_{2}+l_{3}, m_{1}+m_{2}+m_{3}, n_{1}+n_{2}+n_{3}$ makes equal angles with them.

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Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .

If $|\vec{a}|=3, |\vec{b}|=5$ and $|\vec{c}|=7$ and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the angle between $\vec{a}$ and $\vec{b}$ is

Let $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$ 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 $|\overrightarrow{r}|$ is equal to:

Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}$,$\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}$,and $\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ be three vectors such that the projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{a}$. If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$,then $|\vec{b}|$ is equal to

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