If vec{a} = 7hat{j} + 10hat{k}, vec{b} = -hat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - \hat{j} - 4\hat{k}$, so $c = |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{18} = 3\sqrt{2}$.

Vedclass · Accepted Answer

$(2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$

Vedclass · Answer

(A) Let $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - \hat{j} - 4\hat{k}$, so $c = |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{18} = 3\sqrt{2}$.Let $\vec{AC} = \vec{c} - \vec{a} = -4\hat{i} + 2\hat{j} - 4\hat{k}$, so $b = |\vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{36} = 6$.The angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio $c:b = 3\sqrt{2}:6 = 1:\sqrt{2}$.Using the section formula, the position vector $\vec{p}$ of the point dividing $BC$ in ratio $m:n$ is $\frac{m\vec{c} + n\vec{b}}{m+n}$.Here $m=1, n=\sqrt{2}$, so $\vec{p} = \frac{1\vec{c} + \sqrt{2}\vec{b}}{1+\sqrt{2}} = \frac{(-4\hat{i} + 9\hat{j} + 6\hat{k}) + \sqrt{2}(-\hat{i} + 6\hat{j} + 6\hat{k})}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})\hat{i} + (9+6\sqrt{2})\hat{j} + (6+6\sqrt{2})\hat{k}}{1+\sqrt{2}}$.Rationalizing the coefficients: $\frac{-4-\sqrt{2}}{1+\sqrt{2}} = \frac{(-4-\sqrt{2})(\sqrt{2}-1)}{1} = -4\sqrt{2} + 4 - 2 + \sqrt{2} = 2 - 3\sqrt{2}$.$\frac{9+6\sqrt{2}}{1+\sqrt{2}} = \frac{(9+6\sqrt{2})(\sqrt{2}-1)}{1} = 9\sqrt{2} - 9 + 12 - 6\sqrt{2} = 3 + 3\sqrt{2}$.$\frac{6+6\sqrt{2}}{1+\sqrt{2}} = 6$. Thus, $\vec{p} = (2 - 3\sqrt{2})\hat{i} + (3 + 3\sqrt{2})\hat{j} + 6\hat{k}$.

Similar Questions

In the given figure (a square),identify the co-initial vectors.

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

If $\overline{a} = m \overline{b} + n \overline{c}$,where $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,then $m + n =$

If $PQ + QR = (2\lambda^2 - 5)RP$,then $\lambda$ is equal to

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively,in the ratio $2: 1$ internally.

Vedclass Test Series

Exam Paper Generator

Online Exam Module