⇒Vectors

Correct answer option is (B)

Refer figure

Let AD be the angle bisector of angle A which divides BC in the ratio AB : AC.

Here AB = $\sqrt{9+16}=\sqrt{25}$ and

AC = $\sqrt{9+25+16}$

$=\sqrt{50}$

$\therefore$ D divides BC in the ratio $\sqrt{25}:\sqrt{50}$ i.e. 1 : 2

$\therefore$ Position vector of $D=\frac{(4)(2)\hat{k}+5\hat{j}+4\hat{k}}{1+2}=\frac{5j+12\hat{k}}{3}$

Correct answer option is (D)

Refer figure.

$\mathrm{p},\mathrm{q},\mathrm{r},\mathrm{s},\mathrm{m},\mathrm{n}$ be the position vectors of points $\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S},\mathrm{M},\mathrm{N}$.

Now $\overline{\mathrm{PS}}+\overline{\mathrm{QR}}$

$=\bar{\mathrm{s}}-\bar{\mathrm{p}}+\bar{\mathrm{r}}-\bar{\mathrm{q}}=(\bar{\mathrm{s}}+\bar{\mathrm{r}})-(\bar{\mathrm{p}}+\bar{\mathrm{q}})$

Since $M$ and $N$ are mid-points of $PQ$ and RS respectively, we write

$\bar{\mathrm{m}}=\frac{\bar{\mathrm{p}}+\bar{\mathrm{q}}}{2}\text{ and }\bar{\mathrm{n}}=\frac{\bar{\mathrm{r}}+\bar{\mathrm{s}}}{2}$

$\therefore$ eq. (1) becomes

$\overline{\mathrm{PS}}+\overline{\mathrm{QR}}=2\bar{\mathrm{n}}-2\bar{\mathrm{m}}=2(\bar{\mathrm{n}}-\bar{\mathrm{m}})=2\overline{\mathrm{MN}}$

From given data, $t=2$

Correct answer option is (A)

Let $r=x\overline{a}+y\overline{b}$

$\begin{array}{rl}& \therefore -4\hat{\mathrm{i}}-6\hat{\mathrm{j}}-2\hat{\mathrm{k}}=x(-\hat{\mathrm{i}}+4\hat{\mathrm{j}}+3\hat{\mathrm{k}})+y(-8\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ & =(-x-8y)\hat{i}+(4x-y)\hat{j}+(3x+3y)\hat{k}\\ & \therefore -x-8y=-4......\text{(1)}\\ & 4x-y=-6......\text{(2)}\\ & 3\mathrm{x}+3\mathrm{y}=-2......\text{(3)}\\ & \end{array}$

Solving (1) and (2), we get

$y=\frac{2}{3}\text{ and }x=\frac{-4}{3}$