Let \veca=2 \hati+3 \hatj+4 \hatk, \vecb=\hati-2 \hatj-2 \hatk and \vecc=-\hati+4 \hatj+3 \hatk . If \vecd is a vector perpendicular to both \vecb and \vecc , and \veca \cdot \vecd=18 , then |\veca \times \vecd|^2 is equal to :

Vector Algebra PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q31 with Solution

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Correct answer: B

Since $\vec d$ is perpendicular to both $\vec b$ and $\vec c$ , it must be parallel to $\vec b \times \vec c.$ So let $\vec d=\lambda(\vec b\times \vec c).$
Compute $\vec b\times \vec c$ . Given $\vec b=(1,-2,-2), \qquad \vec c=(-1,4,3).$ Then

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix}.$$ Expanding, $$\vec b\times \vec c= \hat i((-2)(3)-(-2)(4))- \hat j((1)(3)-(-2)(-1))+ \hat k((1)(4)-(-2)(-1)).$$ $$=\hat i(-6+8)-\hat j(3-2)+\hat k(4-2)$$ $$=2\hat i-\hat j+2\hat k.$$ Hence, $$\vec d=\lambda(2,-1,2).$$ 3. Use the condition $\vec a\cdot \vec d=18$. Given $$\vec a=(2,3,4).$$ So $$\vec a\cdot \vec d=(2,3,4)\cdot \lambda(2,-1,2)$$ $$=\lambda(4-3+8)=9\lambda.$$ Given this equals $18$, we get $$9\lambda=18 \implies \lambda=2.$$ Therefore, $$\vec d=(4,-2,4).$$ 4. Now compute $|\vec a\times \vec d|^2$. We use the identity $$|\vec a\times \vec d|^2=|\vec a|^2|\vec d|^2-(\vec a\cdot \vec d)^2.$$ First, $$|\vec a|^2=2^2+3^2+4^2=4+9+16=29.$$ Next, $$|\vec d|^2=4^2+(-2)^2+4^2=16+4+16=36.$$ Also, $$\vec a\cdot \vec d=18.$$ So, $$|\vec a\times \vec d|^2=(29)(36)-18^2$$ $$=1044-324=720.$$ 5. Therefore, the required value is $$\boxed{720}.$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Since $\vec d$ is perpendicular to both $\vec b$ and $\vec c$ , it must be parallel to $\vec b \times \vec c.$ So let $\vec d=\lambda(\vec b\times \vec c).$
Compute $\vec b\times \vec c$ . Given $\vec b=(1,-2,-2), \qquad \vec c=(-1,4,3).$ Then

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix}.$$ Expanding, $$\vec b\times \vec c= \hat i((-2)(3)-(-2)(4))- \hat j((1)(3)-(-2)(-1))+ \hat k((1)(4)-(-2)(-1)).$$ $$=\hat i(-6+8)-\hat j(3-2)+\hat k(4-2)$$ $$=2\hat i-\hat j+2\hat k.$$ Hence, $$\vec d=\lambda(2,-1,2).$$ 3. Use the condition $\vec a\cdot \vec d=18$. Given $$\vec a=(2,3,4).$$ So $$\vec a\cdot \vec d=(2,3,4)\cdot \lambda(2,-1,2)$$ $$=\lambda(4-3+8)=9\lambda.$$ Given this equals $18$, we get $$9\lambda=18 \implies \lambda=2.$$ Therefore, $$\vec d=(4,-2,4).$$ 4. Now compute $|\vec a\times \vec d|^2$. We use the identity $$|\vec a\times \vec d|^2=|\vec a|^2|\vec d|^2-(\vec a\cdot \vec d)^2.$$ First, $$|\vec a|^2=2^2+3^2+4^2=4+9+16=29.$$ Next, $$|\vec d|^2=4^2+(-2)^2+4^2=16+4+16=36.$$ Also, $$\vec a\cdot \vec d=18.$$ So, $$|\vec a\times \vec d|^2=(29)(36)-18^2$$ $$=1044-324=720.$$ 5. Therefore, the required value is $$\boxed{720}.$$ So the correct option is **B**.

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