Let \veca=2 \hati+7 \hatj-\hatk, \vecb=3 \hati+5 \hatk and \vecc=\hati-\hatj+2 \hatk . Let \vecd be a vector which is perpendicular to both \veca and \vecb , and \vecc \cdot \vecd=12 . Then (-\hati+\hatj-\hatk) \cdot(\vecc \times \vecd) is equal to :

Vector Algebra PYQ — JEE Main 2023 · 10 Apr · Shift 2 · Q29 with Solution

View written solutionFree

Correct answer: C

Since $\vec d$ is perpendicular to both $\vec a$ and $\vec b$ , it must be parallel to $\vec a \times \vec b.$ So let $\vec d=\lambda(\vec a\times \vec b).$
Compute $\vec a\times \vec b$ . Given $\vec a=(2,7,-1),\qquad \vec b=(3,0,5).$ Therefore,

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 7 & -1\\ 3 & 0 & 5 \end{vmatrix}$$ $$=\hat i(7\cdot 5-(-1)\cdot 0)-\hat j(2\cdot 5-(-1)\cdot 3)+\hat k(2\cdot 0-7\cdot 3)$$ $$=35\hat i-13\hat j-21\hat k.$$ Hence, $$\vec d=\lambda(35,-13,-21).$$ 3. Use the condition $\vec c\cdot \vec d=12$. Given $$\vec c=(1,-1,2).$$ So, $$\vec c\cdot \vec d=\lambda\big((1)(35)+(-1)(-13)+(2)(-21)\big)$$ $$=\lambda(35+13-42)=6\lambda.$$ Given this equals $12$, we get $$6\lambda=12\implies \lambda=2.$$ Thus, $$\vec d=(70,-26,-42).$$ 4. We need to evaluate $$(-\hat i+\hat j-\hat k)\cdot(\vec c\times \vec d).$$ Let $$\vec p=(-1,1,-1).$$ Then this is the scalar triple product $$\vec p\cdot(\vec c\times \vec d).$$ Since $\vec d=2(\vec a\times \vec b)$, $$\vec c\times \vec d=\vec c\times (70,-26,-42).$$ Now compute: $$\vec c\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 2\\ 70 & -26 & -42 \end{vmatrix}$$ $$=\hat i\big((-1)(-42)-2(-26)\big)-\hat j\big((1)(-42)-2(70)\big)+\hat k\big((1)(-26)-(-1)(70)\big)$$ $$=\hat i(42+52)-\hat j(-42-140)+\hat k(-26+70)$$ $$=94\hat i+182\hat j+44\hat k.$$ 5. Dot with $\vec p=(-1,1,-1)$: $$(-1,1,-1)\cdot(94,182,44)=-94+182-44=44.$$ Therefore, $$(-\hat i+\hat j-\hat k)\cdot(\vec c\times \vec d)=44.$$ 6. Comparing with the options, the correct choice is $$\boxed{\text{C: }44}.$$

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View written solutionFree

Correct answer: C

Since $\vec d$ is perpendicular to both $\vec a$ and $\vec b$ , it must be parallel to $\vec a \times \vec b.$ So let $\vec d=\lambda(\vec a\times \vec b).$
Compute $\vec a\times \vec b$ . Given $\vec a=(2,7,-1),\qquad \vec b=(3,0,5).$ Therefore,

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 7 & -1\\ 3 & 0 & 5 \end{vmatrix}$$ $$=\hat i(7\cdot 5-(-1)\cdot 0)-\hat j(2\cdot 5-(-1)\cdot 3)+\hat k(2\cdot 0-7\cdot 3)$$ $$=35\hat i-13\hat j-21\hat k.$$ Hence, $$\vec d=\lambda(35,-13,-21).$$ 3. Use the condition $\vec c\cdot \vec d=12$. Given $$\vec c=(1,-1,2).$$ So, $$\vec c\cdot \vec d=\lambda\big((1)(35)+(-1)(-13)+(2)(-21)\big)$$ $$=\lambda(35+13-42)=6\lambda.$$ Given this equals $12$, we get $$6\lambda=12\implies \lambda=2.$$ Thus, $$\vec d=(70,-26,-42).$$ 4. We need to evaluate $$(-\hat i+\hat j-\hat k)\cdot(\vec c\times \vec d).$$ Let $$\vec p=(-1,1,-1).$$ Then this is the scalar triple product $$\vec p\cdot(\vec c\times \vec d).$$ Since $\vec d=2(\vec a\times \vec b)$, $$\vec c\times \vec d=\vec c\times (70,-26,-42).$$ Now compute: $$\vec c\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 2\\ 70 & -26 & -42 \end{vmatrix}$$ $$=\hat i\big((-1)(-42)-2(-26)\big)-\hat j\big((1)(-42)-2(70)\big)+\hat k\big((1)(-26)-(-1)(70)\big)$$ $$=\hat i(42+52)-\hat j(-42-140)+\hat k(-26+70)$$ $$=94\hat i+182\hat j+44\hat k.$$ 5. Dot with $\vec p=(-1,1,-1)$: $$(-1,1,-1)\cdot(94,182,44)=-94+182-44=44.$$ Therefore, $$(-\hat i+\hat j-\hat k)\cdot(\vec c\times \vec d)=44.$$ 6. Comparing with the options, the correct choice is $$\boxed{\text{C: }44}.$$

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