Question

Let \(\vec{a}\) = 2\(\widehat{i}\)+3\(\widehat{j}\)+4\(\widehat{k}\), \(\vec{b}\) = \(\widehat{i}\)-2\(\widehat{j}\)-2\(\widehat{k}\), \(\vec{c}\) = -\(\widehat{i}\)+4\(\widehat{j}\)+3\(\widehat{k}\) and \(\vec{d}\) is a vector perpendicular to \(\vec{b}\) and \(\vec{c}\),  \(\vec{a}\).\(\vec{d}\) = 18, then find |\(\vec{a}\)x\(\vec{d}\)|²

• A. 720
• B. 700
• C. 360
• D. 300

Answer 1

The Correct Option is A

\(\vec{d}=\lambda (\vec{b}\times \vec{c})=\lambda (2\widehat{i}-\widehat{j}+2\widehat{k})\)
\(\vec{a}.\vec{d}=18\)
\(\Rightarrow \lambda =2\)
Therefore, \(|\vec{a}\times \vec{d}|^{2}=\vec{a}^{2}\vec{d}^{2}-(\vec{a}.\vec{d})^{2}\)
\(\Rightarrow |\vec{a}\times \vec{d}|^{2}=29\times 36-324=1044-324=720\)

Answer 2

Vector Calculations: Cross Product and Magnitude 

Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, $\vec{d}$ must be parallel to the cross product of $\vec{b}$ and $\vec{c}$. So, we can write $\vec{d} = \lambda (\vec{b} \times \vec{c})$ for some scalar $\lambda$.

First, let's calculate $\vec{b} \times \vec{c}$:

$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix} = (-6 + 8)\hat{i} - (3 - 2)\hat{j} + (4 - 2)\hat{k} = 2\hat{i} - \hat{j} + 2\hat{k}$.

So, $\vec{d} = \lambda (2\hat{i} - \hat{j} + 2\hat{k})$. We are given that $\vec{a} \cdot \vec{d} = 18$. Substituting the expressions for $\vec{a}$ and $\vec{d}$:

$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda (2\hat{i} - \hat{j} + 2\hat{k}) = 18$.
$\lambda (4 - 3 + 8) = 18 \Rightarrow 9\lambda = 18 \Rightarrow \lambda = 2$.

Therefore, $\vec{d} = 2(2\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} - 2\hat{j} + 4\hat{k}$.

Now, we need to calculate $|\vec{a} \times \vec{d}|^2$. We know that $|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$.

$|\vec{a}|^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$.
$|\vec{d}|^2 = 4^2 + (-2)^2 + 4^2 = 16 + 4 + 16 = 36$.
$|\vec{a} \times \vec{d}|^2 = (29)(36) - (18)^2 = 1044 - 324 = 720$.