Let three vectors , \overrightarrowa=\alpha \hati+4 \hatj+2 \hatk, \overrightarrowb=5 \hati+3 \hatj+4 \hatk, \overrightarrowc=x \hati+y \hatj+z \hatk form a triangle such that \vecc=\veca-\vecb and the area of the triangle is 5 \sqrt6 . If \alpha is a positive real number, then |\vecc|^2 is equal to:

Vector Algebra PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q32 with Solution

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Correct answer: A

Given vectors

$\vec a=\alpha \hat i+4\hat j+2\hat k,\quad \vec b=5\hat i+3\hat j+4\hat k$

and

$\vec c=\vec a-\vec b.$

So,

$\vec c=(\alpha-5)\hat i+(4-3)\hat j+(2-4)\hat k=(\alpha-5)\hat i+\hat j-2\hat k.$

Hence,

$|\vec c|^2=(\alpha-5)^2+1^2+(-2)^2=(\alpha-5)^2+5.$

Use the area condition

If three vectors $\vec a,\vec b,\vec c$ form a triangle and $\vec c=\vec a-\vec b$ , then the area of the triangle formed by sides $\vec a$ and $\vec b$ is

$\frac12 |\vec a\times \vec b|.$

Given area is $5\sqrt6$ , so

$\frac12 |\vec a\times \vec b|=5\sqrt6$

which gives

$|\vec a\times \vec b|=10\sqrt6.$

Therefore,

$|\vec a\times \vec b|^2=600.$

Compute $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ \alpha & 4 & 2\\ 5 & 3 & 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)$$ $$=10\hat i-(4\alpha-10)\hat j+(3\alpha-20)\hat k.$$ So, $$|\vec a\times \vec b|^2=10^2+(4\alpha-10)^2+(3\alpha-20)^2.$$ Set this equal to $600$: $$100+(4\alpha-10)^2+(3\alpha-20)^2=600.$$ Expand: $$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$$ $$25\alpha(\alpha-8)=0.$$ Thus, $$\alpha=0 \quad \text{or} \quad \alpha=8.$$ Since $\alpha$ is positive, we take $$\alpha=8.$$ --- 4. **Now find $|\vec c|^2$** $$|\vec c|^2=(\alpha-5)^2+5=(8-5)^2+5=9+5=14.$$ --- 5. **Option check** - A: $14$ ✅ - B: $12$ - C: $16$ - D: $10$ So the correct option is **A**.

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View written solutionFree

Correct answer: A

Given vectors

$\vec a=\alpha \hat i+4\hat j+2\hat k,\quad \vec b=5\hat i+3\hat j+4\hat k$

and

$\vec c=\vec a-\vec b.$

So,

$\vec c=(\alpha-5)\hat i+(4-3)\hat j+(2-4)\hat k=(\alpha-5)\hat i+\hat j-2\hat k.$

Hence,

$|\vec c|^2=(\alpha-5)^2+1^2+(-2)^2=(\alpha-5)^2+5.$

Use the area condition

If three vectors $\vec a,\vec b,\vec c$ form a triangle and $\vec c=\vec a-\vec b$ , then the area of the triangle formed by sides $\vec a$ and $\vec b$ is

$\frac12 |\vec a\times \vec b|.$

Given area is $5\sqrt6$ , so

$\frac12 |\vec a\times \vec b|=5\sqrt6$

which gives

$|\vec a\times \vec b|=10\sqrt6.$

Therefore,

$|\vec a\times \vec b|^2=600.$

Compute $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ \alpha & 4 & 2\\ 5 & 3 & 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)$$ $$=10\hat i-(4\alpha-10)\hat j+(3\alpha-20)\hat k.$$ So, $$|\vec a\times \vec b|^2=10^2+(4\alpha-10)^2+(3\alpha-20)^2.$$ Set this equal to $600$: $$100+(4\alpha-10)^2+(3\alpha-20)^2=600.$$ Expand: $$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$$ $$25\alpha(\alpha-8)=0.$$ Thus, $$\alpha=0 \quad \text{or} \quad \alpha=8.$$ Since $\alpha$ is positive, we take $$\alpha=8.$$ --- 4. **Now find $|\vec c|^2$** $$|\vec c|^2=(\alpha-5)^2+5=(8-5)^2+5=9+5=14.$$ --- 5. **Option check** - A: $14$ ✅ - B: $12$ - C: $16$ - D: $10$ So the correct option is **A**.

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