The points whose position vectors are $2\hat{i}+3\hat{j}+4\hat{k}$, $3\hat{i}+4\hat{j}+2\hat{k}$, and $4\hat{i}+2\hat{j}+3\hat{k}$ are the vertices of

Let $OA = a, OB = b$ be two non-collinear vectors, $OP = x_1 a + y_1 b, OQ = x_2 a + y_2 b$ and $A^{\prime}O = OA, B^{\prime}O = OB$. If $x_1 = -\frac{3}{4}, x_2 = \frac{1}{3}, y_1 = \frac{7}{4}, y_2 = \frac{5}{3}$, then

If the vectors $\vec{AB} = -3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$, then the length of the median through $A$ is

The vector having magnitude of $2 \sqrt{29}$ units in the direction of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$ is . . . . . . .

If $\vec{a}$ is a non-zero vector of modulus $|\vec{a}|$ and $m$ is a non-zero scalar,then $m\vec{a}$ is a unit vector if:

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$, $-2\bar{i}+\bar{j}+3\bar{k}$, and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron, then $GD=$