Let A(3 hat{i}+hat{j}-hat{k}) and B(13 hat{i} | vedclass

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

Question

(D) Given, position vectors $\vec{a} = 3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 13 \hat{i}-4 \hat{j}+9 \hat{k}$.The distance $AB = |\vec{b} - \vec{a}

Vedclass · Accepted Answer

$9 \hat{i}-2 \hat{j}+5 \hat{k},-\hat{i}+3 \hat{j}-5 \hat{k}$

Vedclass · Answer

(D) Given, position vectors $\vec{a} = 3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 13 \hat{i}-4 \hat{j}+9 \hat{k}$.The distance $AB = |\vec{b} - \vec{a}| = \sqrt{(13-3)^2 + (-4-1)^2 + (9-(-1))^2} = \sqrt{10^2 + (-5)^2 + 10^2} = \sqrt{100+25+100} = \sqrt{225} = 15$.Since $C$ lies between $A$ and $B$ at a distance of $9$ from $A$, $C$ divides $AB$ in the ratio $AC:CB = 9:(15-9) = 9:6 = 3:2$.Using the section formula, the position vector of $C$ is $\vec{c} = \frac{2\vec{a} + 3\vec{b}}{3+2} = \frac{2(3 \hat{i}+\hat{j}-\hat{k}) + 3(13 \hat{i}-4 \hat{j}+9 \hat{k})}{5} = \frac{(6+39) \hat{i} + (2-12) \hat{j} + (-2+27) \hat{k}}{5} = \frac{45 \hat{i} - 10 \hat{j} + 25 \hat{k}}{5} = 9 \hat{i}-2 \hat{j}+5 \hat{k}$.$D$ lies on the line $L$ on the other side of $A$ at a distance of $6$ units. Thus, $A$ is the midpoint of $DC$ if we consider the segment $DC$ where $AD=6$ and $AC=9$ is not correct; rather $A$ divides $DC$ externally. Since $D$ is on the other side of $A$, $A$ is between $D$ and $C$. The ratio $DA:AC = 6:9 = 2:3$. Thus $A$ divides $DC$ in the ratio $2:3$ internally.$\vec{a} = \frac{3\vec{d} + 2\vec{c}}{3+2} \Rightarrow 5\vec{a} = 3\vec{d} + 2\vec{c} \Rightarrow 3\vec{d} = 5\vec{a} - 2\vec{c}$.$3\vec{d} = 5(3 \hat{i}+\hat{j}-\hat{k}) - 2(9 \hat{i}-2 \hat{j}+5 \hat{k}) = (15-18) \hat{i} + (5+4) \hat{j} + (-5-10) \hat{k} = -3 \hat{i} + 9 \hat{j} - 15 \hat{k}$.$\vec{d} = -\hat{i} + 3 \hat{j} - 5 \hat{k}$.

Let $A(3 \hat{i}+\hat{j}-\hat{k})$ and $B(13 \hat{i}-4 \hat{j}+9 \hat{k})$ be two points on a line $L$. $C$ and $D$ are points on $L$ on either side of $A$ at distances of $9$ and $6$ units respectively, and $C$ lies between $A$ and $B$. Then the position vectors of $C$ and $D$ are respectively:

Similar Questions

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$,then find $\theta$ and the components of $\vec{a}$.

The projection of the vector $i+j+k$ along the vector $j$ is

Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point $(-5,7).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module