Vector Algebra

The distance of the point having position vector $- \widehat i + 2\widehat j + 6\widehat k$ from the straight line passing through the point (2, 3, – 4) and parallel to the vector, $6\widehat i + 3\widehat j - 4\widehat k$ is :

Let $\overrightarrow \alpha = 3\widehat i + \widehat j$ and $\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$ . If $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}$, where ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$ and $\overrightarrow {{\beta _2}}$ is perpendicular to $\overrightarrow \alpha$ , then ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}}$ is equal to

Let the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle be $2 \hat{i}+2 \hat{j}+\hat{k}, \hat{i}+2 \hat{j}+2 \hat{k}$ and $2 \hat{i}+\hat{j}+2 \hat{k}$ respectively. Let $l_1, l_2$ and $l_3$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ respectively, then $l_1^2+l_2^2+l_3^2$ equals:

Let $\overrightarrow a$ , $\overrightarrow b$ and $\overrightarrow c$ be three unit vectors such that $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$. If $\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a$ and $\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a$, then the ordered pair, $\left( {\lambda ,\overrightarrow d } \right)$ is equal to :

Let $\overrightarrow a = \widehat i - 2\widehat j + \widehat k$ and $\overrightarrow b = \widehat i - \widehat j + \widehat k$ be two vectors. If $\overrightarrow c$ is a vector such that $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$ and $\overrightarrow c .\overrightarrow a = 0$, then $\overrightarrow c .\overrightarrow b$ is equal to

Let $\overrightarrow a$ = $\widehat i$ + 2$\widehat j$ $-$ 3$\widehat k$ and $\overrightarrow b = 2\widehat i$ $-$ 3$\widehat j$ + 5$\widehat k$. If $\overrightarrow r$ $\times$ $\overrightarrow a$ = $\overrightarrow b$ $\times$ $\overrightarrow r$, $\overrightarrow r$ . $\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$ = 3 and $\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$ = $-$1, $\alpha$ $\in$ R, then the value of $\alpha$ + ${\left| {\overrightarrow r } \right|^2}$ is equal to :

Let $\overrightarrow a$ and $\overrightarrow b$ be two non-zero vectors perpendicular to each other and $|\overrightarrow a | = |\overrightarrow b |$. If $|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$, then the angle between the vectors $\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$ and ${\overrightarrow a }$ is equal to :

If $\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$ and $\left| {\overrightarrow a \times \overrightarrow b } \right|$ = 8, then $\left| {\overrightarrow a .\,\overrightarrow b } \right|$ is equal to :

Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ and $\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$. Then the vector product $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$ is equal to :

Let $\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$, $\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$ and $\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$ where $\alpha ,\,\beta \in R$, be three vectors. If the projection of $\overrightarrow a$ on $\overrightarrow c$ is ${{10} \over 3}$ and $\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$, then the value of $\alpha + \beta$ is equal to :