Question

The vertices of a quadrialteral are $A(1,2,-1),B(-4,2,-2),C(4,1,-5)$ and $D(2,-1,3)$. Forces of magnitudes $2N,3N,2N$ are acting at point $A$ along the lines $AB,AC$ and $AD$ respectively. Their resultant is:

• A. $\frac{10\hat{i}-9\hat{j}+6\hat{k}}{\sqrt{26}}$
• B. $\left(\frac{\hat{i}-9\hat{j}-6\hat{k}}{\sqrt{26}}\right)$
• C. $\frac{\hat{i}+9\hat{j}+16\hat{k}}{\sqrt{26}}$
• D. $\frac{\hat{i}-19\hat{j}+6\hat{k}}{\sqrt{26}}$

Answer 1

The correct option is B $\left(\frac{\hat{i}-9\hat{j}-6\hat{k}}{\sqrt{26}}\right)$

$\overrightarrow{AB}\equiv (-4,2,-2)-(1,2,-1)\equiv (-5,0,-1)$
$AB=\sqrt{5^2+1^2}=\sqrt{26}$
$\widehat{AB}=\frac{-5\hat{i}-\hat{k}}{\sqrt{26}}$
$\overrightarrow{AC}\equiv (4,1,-5)-(1,2,-1)\equiv (3,-1,-4)$
$AC=\sqrt{3^2+1^2+4^2}=\sqrt{26}$
$\widehat{AC}=\frac{3\hat{i}-\hat{j}-4\hat{k}}{\sqrt{26}}$
$\overrightarrow{AD}\equiv (2,-1,3)-(1,2,-1)\equiv (1,-3,4)$
$AD\equiv \sqrt{1^2+3^2+4^2}$
$=\sqrt{26}$
$\widehat{AD}=\frac{\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{26}}$
Now
${\overrightarrow{F}}_1=2\widehat{AB}=\frac{-10\hat{i}-2\hat{k}}{\sqrt{26}}$
${\overrightarrow{F}}_2=3\widehat{AC}=\frac{9\hat{i}-3\hat{j}-12\hat{k}}{\sqrt{26}}$
${\overrightarrow{F}}_3=2\widehat{AD}=\frac{2\hat{i}-6\hat{j}+8\hat{k}}{\sqrt{26}}$
${\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=\frac{\hat{i}-9\hat{j}-6\hat{k}}{\sqrt{26}}$