Let three vectors , a= i+4j+2k, b=5i+3j+4k, c=xi+yj+zk form a triangle such that c=a-b and…

Question

Let three vectors ,$\overset{	o }{a}=\alpha \hat{i}+4\hat{j}+2\hat{k},\overset{	o }{b}=5\hat{i}+3\hat{j}+4\hat{k},\overset{	o }{c}=x\hat{i}+y\hat{j}+z\hat{k}$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5\sqrt{6}$. If $\alpha $ is a positive real number, then $|\vec{c}{|}^{2}$ is equal to:

Accepted Answer

To solve this, let's start with the given vector equation:

$\vec{c}=\vec{a}-\vec{b}$

Given vectors are:

$\overset{	o }{a}=\alpha \hat{i}+4\hat{j}+2\hat{k}$

$\overset{	o }{b}=5\hat{i}+3\hat{j}+4\hat{k}$

Then, the vector $\overset{	o }{c}$ is:

$\overset{	o }{c}=(\alpha \hat{i}+4\hat{j}+2\hat{k})-(5\hat{i}+3\hat{j}+4\hat{k})$

$\overset{	o }{c}=(\alpha -5)\hat{i}+(4-3)\hat{j}+(2-4)\hat{k}$

$\overset{	o }{c}=(\alpha -5)\hat{i}+1\hat{j}-2\hat{k}$

The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of the cross product of $\vec{a}$ and $\vec{b}$, divided by 2:

$=\frac{1}{2}|\vec{a}	imes \vec{b}|=5\sqrt{6}$

This implies:

$|\vec{a}	imes \vec{b}|=10\sqrt{6}$

Let's find $\vec{a}	imes \vec{b}$:

$\vec{a}	imes \vec{b}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ \alpha  \quad  4 \quad  2 \ 5 \quad  3 \quad  4 \end{vmatrix}=\hat{i}(4\cdot 4-2\cdot 3)-\hat{j}(\alpha \cdot 4-2\cdot 5)+\hat{k}(\alpha \cdot 3-4\cdot 5)$

$\vec{a}	imes \vec{b}=\hat{i}(16-6)-\hat{j}(4\alpha -10)+\hat{k}(3\alpha -20)$

$\vec{a}	imes \vec{b}=\hat{i}(10)-\hat{j}(4\alpha -10)+\hat{k}(3\alpha -20)$

Magnitude of $\vec{a}	imes \vec{b}$:

$|\vec{a}	imes \vec{b}|=\sqrt{{10}^{2}+(4\alpha -10{)}^{2}+(3\alpha -20{)}^{2}}$

We know:

$\sqrt{{10}^{2}+(4\alpha -10{)}^{2}+(3\alpha -20{)}^{2}}=10\sqrt{6}$

Squaring both sides:

$100+(4\alpha -10{)}^{2}+(3\alpha -20{)}^{2}=600$

$100+16{\alpha }^{2}-80\alpha +100+9{\alpha }^{2}-120\alpha +400=600$

$25{\alpha }^{2}-200\alpha +600=600$

$25{\alpha }^{2}-200\alpha =0$

${\alpha }^{2}-8\alpha =0$

$\alpha (\alpha -8)=0$

Since $\alpha $ is a positive real number:

$\alpha =8$

Then, the vector $\vec{c}$ is:

$\vec{c}=(8-5)\hat{i}+1\hat{j}-2\hat{k}$

$\vec{c}=3\hat{i}+1\hat{j}-2\hat{k}$

Magnitude squared of $\vec{c}$ is:

$|\vec{c}{|}^{2}={3}^{2}+{1}^{2}+(-2{)}^{2}$

$|\vec{c}{|}^{2}=9+1+4$

$|\vec{c}{|}^{2}=14$

Therefore, the correct option is:

Option A: 14

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