Let a=4i-j+3k, b=10i+2j-k and a vector c be such that 2( a b)+3( b c)= 0.

Question

Let $\overset{	o }{a}=4\hat{i}-\hat{j}+3\hat{k},\overset{	o \leftarrow }{b}=10\hat{i}+2\hat{j}-\hat{k}$ and a vector $\overset{	o }{c}$ be such that $2(\overset{	o }{a}	imes \overset{	o }{b})+3(\overset{	o \leftarrow }{b}	imes \overset{	o }{c})=\overset{	o }{0}$.

If $\overset{	o }{a}\cdot \overset{	o }{c}=15$, then $\overset{	o }{c}\cdot (\hat{i}+\hat{j}-3\hat{k})$ is equal to :

Accepted Answer

The unknown vector $\overset{	o }{c}$ is represented in 3D: $\overset{	o }{c}=x\hat{i}+y\hat{j}+z\hat{k}$

Cross product of 2 vectors $\overset{	o }{a}$ and $\overset{	o }{b}$ :

$(\overset{	o }{a}	imes \overset{	o }{b})=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  -1 \quad  3 \ 10 \quad  2 \quad  -1 \end{vmatrix}=-5\hat{i}+34\hat{j}+18\hat{k}\ldots ..(i)$

Cross product of 2 vectors $\overset{	o }{b}$ and $\overset{	o }{c}$ :

$(\overset{	o }{b}	imes \overset{	o }{c})=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 10 \quad  2 \quad  -1 \ x \quad  y \quad  z \end{vmatrix}=(2z+y)\hat{i}-(10z+x)\hat{j}+(10y-2x)\hat{k}\ldots \ldots (ii)$

We know, $2(a	imes b)+3(b	imes c)=0$

$\Rightarrow 2(-5\hat{i}+34\hat{j}+18\hat{k})+3{(2z+y)\hat{i}-(10z+x)\hat{j}+(10y-2x)\hat{k}}=0$

$\Rightarrow$ $(6z+3y-10)\hat{i}+(68-30z-3x)j+(36+30y-6x)\hat{k}=0$

As individual vector components of a null vector are zero.

$\begin{gathered} x-5y =6\ldots .{P}_{1} \ y+2z =\frac{10}{3}\ldots ..{P}_{2} \ 10z+x =\frac{68}{3}\ldots .{P}_{3} \end{gathered}$

$(\vec{a})\cdot (\vec{c})=15\Rightarrow (4\hat{i}-\hat{j}+3\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=4x-y+3z=15\ldots \ldots (iv)$

Equation set $(iii)$ is mathematically constructed to form a linear combination of 2 planes ${P}_{1}$ and ${P}_{2}$ generating the ${3}^{}$ plane ${P}_{3}$ :

$x-5y+(\lambda )\cdot (y+2z)=\frac{68}{3}\Rightarrow \lambda =5\Rightarrow {P}_{1}+5{P}_{2}={P}_{3}\ldots \ldots (v)$

Equation ( $v$ ) is conclusive in generating a family of planes ${{P}_{1},{P}_{2},{P}_{3}}$ indicating infinite solutions for the system equation (iii)

We need an auxiliary equation provided through (iv) to render an unique solution for the vector $\vec{c}$ :

$[[\begin{vmatrix} x-5y \quad  =6 \quad  \ldots . \quad  {P}_{1} \ y+2z \quad  =\frac{10}{3} \quad  \ldots . \quad  {P}_{2} \ 4x-y+3z \quad  =15 \quad  \ldots . \quad  {P}_{4} \end{gathered}]]|\ldots .(vi)$

Equation (vi) can be solved using Determinants and Cramer's rule :

$\Delta =|\begin{gathered} 1 \quad  -5 \quad  0 \ 0 \quad  1 \quad  2 \ 4 \quad  -1 \quad  3 \end{vmatrix}\, \leftarrow \, {R}_{3}={R}_{3}-4{R}_{1}=\begin{vmatrix} 1 \quad  -5 \quad  0 \ 0 \quad  1 \quad  2 \ 0 \quad  19 \quad  3 \end{vmatrix}=-35\, \ldots ..(vii)$

${\Delta}_{x}=\begin{vmatrix} 6 \quad  -5 \quad  0 \ \frac{10}{3} \quad  1 \quad  2 \ 15 \quad  -1 \quad  3 \end{vmatrix}$

$\begin{gathered} {R}_{1}={R}_{1}+5{R}_{2} \ {R}_{3}={R}_{3}+{R}_{2} \end{gathered}$

$=\begin{vmatrix} \frac{68}{3} \quad  0 \quad  10 \ \frac{10}{3} \quad  1 \quad  2 \ \frac{55}{3} \quad  0 \quad  5 \end{vmatrix}$

${C}_{1}{C}_{2}$

$=-|\begin{array}{cc} 0 & \frac{68}{3} \ 1 & \frac{10}{3} \ 0 & \frac{10}{3} \ \frac{55}{3} \end{array}|=(\frac{68}{3})\cdot (5)-(\frac{55}{3})\cdot (10)=-70\ldots \ldots (viii)$

Equations $(vii),(viii),(ix)$ and $(x)$ provides the solution to system equation $(vi)$ from Cramer's rule :

$[[|\begin{gathered} x=\frac{{\Delta}_{x}}{\Delta}=\frac{70}{35}= \ y=\frac{{\Delta}_{y}}{\Delta}=-\frac{28}{35}=-\frac{4}{5} \ z=\frac{{\Delta}_{z}}{\Delta}=\frac{217}{(3)\cdot (35)}=\frac{31}{15} \end{gathered}]]|$

$\Rightarrow \overset{	o }{c}=(2)\hat{i}-(\frac{4}{5})\hat{j}+(\frac{31}{15})\hat{k}$

$\begin{gathered} 	herefore (\overset{	o }{c})\cdot (\hat{i}+\hat{j}-3\hat{k}) \ ={(2)\hat{i}-(\frac{4}{5})\hat{j}+(\frac{31}{15})\hat{k}}\cdot (\hat{i}+\hat{j}-3\hat{k}) \ =2-\frac{4}{5}-\frac{31}{5} \end{gathered}$

$=-\frac{4+31-10}{5}=-5$

Solutions · 1