If $a, b, c$ are distinct real numbers and $P, Q, R$ are three points whose position vectors are respectively $a \hat{i}+b \hat{j}+c \hat{k}$, $b \hat{i}+c \hat{j}+a \hat{k}$ and $c \hat{i}+a \hat{j}+b \hat{k}$, then $\angle Q P R=$

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other,then $\lambda$ is equal to

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a},$ for any two nonzero vectors $\vec{a}$ and $\vec{b}.$

Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. The quadrilateral $PQRS$ must be a

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

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