Question

The vertices of a quadrilateral are $A(1,2,-1),B(-4,2,-2),C(4,1,-5)$ and $D(2,-1,3).$ At the point $A$ forces of magnitudes $2,3,2$ kg act along $AB,AC$ and $AD$ respectively. Find their resultant.

• A. $\frac{2}{\sqrt{\left(26\right)}},\frac{-7}{\sqrt{\left(26\right)}},\frac{-1}{\sqrt{\left(26\right)}}$
• B. $\frac{-2}{\sqrt{\left(26\right)}},\frac{7}{\sqrt{\left(26\right)}},\frac{1}{\sqrt{\left(26\right)}}$
• C. $\frac{1}{\sqrt{\left(118\right)}},\frac{-9}{\sqrt{\left(118\right)}},\frac{-6}{\sqrt{\left(118\right)}}$
• D. $\frac{-1}{\sqrt{\left(118\right)}},\frac{9}{\sqrt{\left(118\right)}},\frac{6}{\sqrt{\left(118\right)}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{\left(118\right)}},\frac{-9}{\sqrt{\left(118\right)}},\frac{-6}{\sqrt{\left(118\right)}}$

In terms of unit vectors $A(1,2,-1)$
is $i+2j-k$ and similarly we can write down the other
points with respect to O as origin.
$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(-4i+2j-2k)-(i+2j-k)$ or $\overrightarrow{AB}=-5i+0j-k\therefore \left|AB\right|=\sqrt{25+1}=\sqrt{26}.$
$\therefore$ Unit vector along $AB=\frac{1}{\sqrt{26}}(-5i-k).$
$\therefore$ A force of magnitude 2 kg along AB is $2\cdot \frac{1}{\sqrt{\left(26\right)}}(-5i-k)=\frac{1}{\sqrt{\left(26\right)}}(-10i-2k)...\left(1\right)$
Similarly forces of magnitudes 3 and 2 kg along AC and AD are respectively
$3\cdot \frac{1}{\sqrt{\left(26\right)}}(3i-j-4k)=\frac{1}{\sqrt{\left(26\right)}}(9i-3j-12k)...\left(2\right)$ and
$2\cdot \frac{1}{\sqrt{\left(26\right)}}(i-3j+4k)=\frac{1}{\sqrt{\left(26\right)}}(-2i-6j+8k)...\left(3\right)$
Adding (1), (2) and (3), the resultant R is $R=\frac{1}{\sqrt{\left(26\right)}}(i-9j-6k)$
Its magnitude is $\sqrt{\left(\frac{1+81+36}{26}\right)}=\sqrt{\left(\frac{118}{26}\right)}kg.$
Also d.c.'s of the resultant are
$\frac{1}{\sqrt{\left(118\right)}},\frac{-9}{\sqrt{\left(118\right)}},\frac{-6}{\sqrt{\left(118\right)}}$