Let $|\vec a|=2$, $|\vec b|=3$ and the angle between the vectors $\vec a$ and $\vec b$ be $\dfrac{\pi}{4}$. Then $|(\vec a+2\vec b)\times(2\vec a-3\vec b)|^2$ is equal to:
Let $\vec a=3\hat i+\hat j-2\hat k,\ \vec b=4\hat i+\hat j+7\hat k$ and $\vec c=\hat i-3\hat j+4\hat k$ be three vectors.
If a vector $\vec p$ satisfies $\vec p\times\vec b=\vec c\times\vec b$ and $\vec p\cdot\vec a=0$,
then $\vec p\cdot(\hat i-\hat j-\hat k)$ is equal to:
The distance of the point $Q(0,2,-2)$ from the line passing through the point
$P(5,-4,3)$ and perpendicular to the lines
$\ \vec r = (-3\hat i + 2\hat k) + \lambda(2\hat i + 3\hat j + 5\hat k),\ \lambda\in\mathbb R,$
and
$\ \vec r = (\hat i - 2\hat j + \hat k) + \mu(-\hat i + 3\hat j + 2\hat k),\ \mu\in\mathbb R,$
is:
Let $\overrightarrow a $ and $\overrightarrow b $ be two vectors such that $\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$ and the angle between $\overrightarrow a $ and $\overrightarrow b $ is 60$^\circ$. If ${1 \over 8}\overrightarrow a $ is a unit vector, then $\left| {\overrightarrow b } \right|$ is equal to :
Let $\overrightarrow{OA}=2\vec a,\ \overrightarrow{OB}=6\vec a+5\vec b,\ \overrightarrow{OC}=3\vec b$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units, then the area (in sq. units) of the quadrilateral $OABC$ is equal to:
Let $\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$, where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow a $ and $\overrightarrow b $ is $\sqrt {15({\alpha ^2} + 4)} $, then the value of $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a \,.\,\overrightarrow b } \right){\left| {\overrightarrow b } \right|^2}$ is equal to :
If $\vec{a}$ is a nonzero vector such that its projections on the vectors
$2\hat{i} - \hat{j} + 2\hat{k}$, $\hat{i} + 2\hat{j} - 2\hat{k}$ and $\hat{k}$ are equal,
then a unit vector along $\vec{a}$ is:
For a triangle $ABC$,
$\overrightarrow{AB}=-2\hat i+\hat j+3\hat k$
$\overrightarrow{CB}=\alpha\hat i+\beta\hat j+\gamma\hat k$
$\overrightarrow{CA}=4\hat i+3\hat j+\delta\hat k$
If $\delta>0$ and the area of the triangle $ABC$ is $5\sqrt{6}$, then $\overrightarrow{CB}\cdot\overrightarrow{CA}$ is equal to:
Let a, b c $ \in $ R be such that a2 + b2 + c2 = 1. If $a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$, where${\theta = {\pi \over 9}}$, then the angle between the vectors $a\widehat i + b\widehat j + c\widehat k$ and $b\widehat i + c\widehat j + a\widehat k$ is
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 :