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

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(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

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

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(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:

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