Let \overrightarrowa=6 \hati+\hatj-\hatk and \overrightarrowb=\hati+\hatj . If \overrightarrowc is a is vector such that |\overrightarrowc| \geq 6, \overrightarrowa \cdot \overrightarrowc=6|\overrightarrowc|,|\overrightarrowc-\overrightarrowa|=2 \sqrt2 and the angle between \veca \times \vecb and \vecc is 60^\circ , then |(\veca \times \vecb) \times \vecc| is equal to:

Vector Algebra PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q38 with Solution

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Correct answer: C

Given vectors

$\vec a=6\hat i+\hat j-\hat k=(6,1,-1), \qquad \vec b=\hat i+\hat j=(1,1,0)$

We need to find

$|(\vec a\times \vec b)\times \vec c|$

given:

$|\vec c|\ge 6$
$\vec a\cdot \vec c=6|\vec c|$
$|\vec c-\vec a|=2\sqrt 2$
angle between $\vec a\times \vec b$ and $\vec c$ is $60^\circ$

Compute $|\vec a|$

$|\vec a|=\sqrt{6^2+1^2+(-1)^2}=\sqrt{38}$

Also,

So,

$\cos\theta=\frac{6}{\sqrt{38}}$

where $\theta$ is the angle between $\vec a$ and $\vec c$ .

Use $|\vec c-\vec a|=2\sqrt 2$

We know

$|\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2-2\vec a\cdot \vec c$

Substitute values:

$8=|\vec c|^2+38-2(6|\vec c|)$

$8=|\vec c|^2+38-12|\vec c|$

$|\vec c|^2-12|\vec c|+30=0$

Solving,

$|\vec c|=\frac{12\pm\sqrt{144-120}}{2}=\frac{12\pm\sqrt{24}}{2}=6\pm \sqrt 6$

Given $|\vec c|\ge 6$ , we get

$|\vec c|=6+\sqrt 6$

Compute $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 6 & 1 & -1\\ 1 & 1 & 0 \end{vmatrix}$$ $$=\hat i(1\cdot 0-(-1)\cdot 1)-\hat j(6\cdot 0-(-1)\cdot 1)+\hat k(6\cdot 1-1\cdot 1)$$ $$=\hat i-\hat j+5\hat k$$ Hence, $$\vec a\times \vec b=(1,-1,5)$$ and $$|\vec a\times \vec b|=\sqrt{1^2+(-1)^2+5^2}=\sqrt{27}=3\sqrt 3$$ --- 5. **Use the angle condition** The angle between $\vec a\times \vec b$ and $\vec c$ is $60^\circ$. Now, $$|(\vec a\times \vec b)\times \vec c|=|\vec a\times \vec b|\,|\vec c|\sin 60^\circ$$ So, $$|(\vec a\times \vec b)\times \vec c|=(3\sqrt 3)(6+\sqrt 6)\cdot \frac{\sqrt 3}{2}$$ $$=\frac{3\cdot 3}{2}(6+\sqrt 6)$$ $$=\frac{9}{2}(6+\sqrt 6)$$ --- 6. **Match with options** This is exactly **Option C**. $$\boxed{\frac{9}{2}(6+\sqrt 6)}$$

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View written solutionFree

Correct answer: C

Given vectors

$\vec a=6\hat i+\hat j-\hat k=(6,1,-1), \qquad \vec b=\hat i+\hat j=(1,1,0)$

We need to find

$|(\vec a\times \vec b)\times \vec c|$

given:

$|\vec c|\ge 6$
$\vec a\cdot \vec c=6|\vec c|$
$|\vec c-\vec a|=2\sqrt 2$
angle between $\vec a\times \vec b$ and $\vec c$ is $60^\circ$

Compute $|\vec a|$

$|\vec a|=\sqrt{6^2+1^2+(-1)^2}=\sqrt{38}$

Also,

So,

$\cos\theta=\frac{6}{\sqrt{38}}$

where $\theta$ is the angle between $\vec a$ and $\vec c$ .

Use $|\vec c-\vec a|=2\sqrt 2$

We know

$|\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2-2\vec a\cdot \vec c$

Substitute values:

$8=|\vec c|^2+38-2(6|\vec c|)$

$8=|\vec c|^2+38-12|\vec c|$

$|\vec c|^2-12|\vec c|+30=0$

Solving,

$|\vec c|=\frac{12\pm\sqrt{144-120}}{2}=\frac{12\pm\sqrt{24}}{2}=6\pm \sqrt 6$

Given $|\vec c|\ge 6$ , we get

$|\vec c|=6+\sqrt 6$

Compute $\vec a\times \vec b$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 6 & 1 & -1\\ 1 & 1 & 0 \end{vmatrix}$$ $$=\hat i(1\cdot 0-(-1)\cdot 1)-\hat j(6\cdot 0-(-1)\cdot 1)+\hat k(6\cdot 1-1\cdot 1)$$ $$=\hat i-\hat j+5\hat k$$ Hence, $$\vec a\times \vec b=(1,-1,5)$$ and $$|\vec a\times \vec b|=\sqrt{1^2+(-1)^2+5^2}=\sqrt{27}=3\sqrt 3$$ --- 5. **Use the angle condition** The angle between $\vec a\times \vec b$ and $\vec c$ is $60^\circ$. Now, $$|(\vec a\times \vec b)\times \vec c|=|\vec a\times \vec b|\,|\vec c|\sin 60^\circ$$ So, $$|(\vec a\times \vec b)\times \vec c|=(3\sqrt 3)(6+\sqrt 6)\cdot \frac{\sqrt 3}{2}$$ $$=\frac{3\cdot 3}{2}(6+\sqrt 6)$$ $$=\frac{9}{2}(6+\sqrt 6)$$ --- 6. **Match with options** This is exactly **Option C**. $$\boxed{\frac{9}{2}(6+\sqrt 6)}$$

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