Let $\vec{\alpha}=4\hat{i}+3\hat{j}+5\hat{k}$ and $\vec{\beta}=\hat{i}+2\hat{j}-4\hat{k}$.
Let $\vec{\beta}_{1}$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ be perpendicular to $\vec{\alpha}$.
If $\vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}$, then the value of $5\,\vec{\beta}_{2}\cdot(\hat{i}+\hat{j}+\hat{k})$ is:
Let $\vec a,\vec b,\vec c$ be three non-zero vectors such that $\vec b$ and $\vec c$ are non-collinear. If $\ \vec a+5\vec b\ $ is collinear with $\vec c$, and $\ \vec b+6\vec c\ $ is collinear with $\vec a$, and $\ \vec a+\alpha,\vec b+\beta,\vec c=\vec 0$, then $\alpha+\beta$ is equal to:
Let $\vec a=\hat i+2\hat j+\hat k$ and $\vec b=2\hat i+\hat j-\hat k$. Let $\vec c$ be a unit vector in the plane of the vectors $\vec a$ and $\vec b$ and be perpendicular to $\vec a$. Then such a vector $\vec c$ is:
Let $A(2,3,5)$ and $C(-3,4,-2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD}= \hat{i}+2\hat{j}+3\hat{k}$, then the area of the parallelogram is equal to:
Let $\overrightarrow a = \widehat i + \widehat j - \widehat k$ and $\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$. Then the number of vectors $\overrightarrow b $ such that $\overrightarrow b \times \overrightarrow c = \overrightarrow a $ and $|\overrightarrow b | \in $ {1, 2, ........, 10} is :
$L_1:;\vec r=(2+\lambda),\hat i+(1-3\lambda),\hat j+(3+4\lambda),\hat k,;\lambda\in\mathbb R$
$L_2:;\vec r=2(1+\mu),\hat i+3(1+\mu),\hat j+(5+\mu),\hat k,;\mu\in\mathbb R$
is $\dfrac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals
Let $O$ be the origin and the position vectors of $A$ and $B$ be
$\vec{A} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ respectively.
If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is:
The vector $\vec{a}=-\hat{i}+2\hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec{b}$.
Then the projection of $3\vec{a}+\sqrt{2}\,\vec{b}$ on $\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}$ is:
Let $\vec{a}=4\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=11\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times\vec{c}=\vec{c}\times(-2\vec{a}+3\vec{b})$. If $(2\vec{a}+3\vec{b})\cdot\vec{c}=1670$, then $|\vec{c}|^{2}$ is equal to: