If 4i + 7j + 8k,2i + 3j + 4k and 2i + 5j + 7k | vedclass

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

Question

(A) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4i + 7j + 8k$,$\vec{b} = 2i + 3j + 4k$,and $\vec{c} = 2i + 5j + 7k$ respectivel

Vedclass · Accepted Answer

$\frac{1}{3}(6i + 13j + 18k)$

Vedclass · Answer

(A) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4i + 7j + 8k$,$\vec{b} = 2i + 3j + 4k$,and $\vec{c} = 2i + 5j + 7k$ respectively.According to the Angle Bisector Theorem,the bisector of angle $A$ divides the opposite side $BC$ in the ratio $AB:AC$.First,we calculate the lengths of sides $AB$ and $AC$:$\vec{AB} = \vec{b} - \vec{a} = (2-4)i + (3-7)j + (4-8)k = -2i - 4j - 4k$.$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$\vec{AC} = \vec{c} - \vec{a} = (2-4)i + (5-7)j + (7-8)k = -2i - 2j - k$.$|\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Let $D$ be the point where the bisector of angle $A$ meets $BC$. The ratio $BD:DC = |\vec{AB}|:|\vec{AC}| = 6:3 = 2:1$.Using the section formula,the position vector of $D$ is given by $\vec{d} = \frac{1(\vec{b}) + 2(\vec{c})}{1+2} = \frac{(2i + 3j + 4k) + 2(2i + 5j + 7k)}{3}$.$\vec{d} = \frac{2i + 3j + 4k + 4i + 10j + 14k}{3} = \frac{6i + 13j + 18k}{3} = \frac{1}{3}(6i + 13j + 18k)$.

If $4i + 7j + 8k$,$2i + 3j + 4k$ and $2i + 5j + 7k$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

Similar Questions

$A$ vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $\bar{a}=\frac{1}{3}(\bar{i}-2 \bar{j}+2 \bar{k})$,$\bar{b}=\frac{1}{5}(-4 \bar{i}-3 \bar{k})$ and $\bar{c}=\bar{j}$,is

The orthogonal projection of vector $\vec{a}$ on vector $\vec{b}$ is:

For all real $x$,the vectors $Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$ make an obtuse angle with each other. Then the value of $C$ can be in:

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

If $a = 2i + 2j + 3k$,$b = -i + 2j + k$,and $c = 3i + j$,then $a + tb$ is perpendicular to $c$ if $t = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module