Question

Let $\vec{a}=\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$, $\vec{c}=6\hat{i}+3\hat{j}-2\hat{k}$ be three vectors. If $\vec{d}$ is a vector perpendicular to both $\vec{a}$, $\vec{b}$ and $|\vec{d}\times\vec{c}|=14$, then $|\vec{d}\cdot\vec{c}|=$

• A. 35
• B. 70
• C. 140
• D. 105

Hint:

A vector perpendicular to two given vectors $\vec{a}$ and $\vec{b}$ is always of the form $\lambda(\vec{a} \times \vec{b})$. The scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ gives the volume of the parallelepiped formed by the three vectors.

Answer 1

The Correct Option is B

Since vector $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, it must be parallel to their cross product, $\vec{a} \times \vec{b}$.
So, $\vec{d} = \lambda (\vec{a} \times \vec{b})$ for some scalar $\lambda$.
First, calculate $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & -2 & -2 \end{vmatrix} = \hat{i}(2 - (-2)) - \hat{j}(-2 - 1) + \hat{k}(-2 - (-1)) = 4\hat{i} + 3\hat{j} - \hat{k}$.
So, $\vec{d} = \lambda(4\hat{i} + 3\hat{j} - \hat{k})$.
Next, we use the condition $|\vec{d}\times\vec{c}|=14$.
$\vec{d}\times\vec{c} = \lambda ((\vec{a} \times \vec{b}) \times \vec{c})$.
Let's calculate the cross product $(4\hat{i} + 3\hat{j} - \hat{k}) \times (6\hat{i} + 3\hat{j} - 2\hat{k})$:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & -1 \\ 6 & 3 & -2 \end{vmatrix} = \hat{i}(-6 - (-3)) - \hat{j}(-8 - (-6)) + \hat{k}(12 - 18) = -3\hat{i} + 2\hat{j} - 6\hat{k}$.
So, $\vec{d}\times\vec{c} = \lambda(-3\hat{i} + 2\hat{j} - 6\hat{k})$.
$|\vec{d}\times\vec{c}| = |\lambda| \sqrt{(-3)^2+2^2+(-6)^2} = |\lambda|\sqrt{9+4+36} = |\lambda|\sqrt{49} = 7|\lambda|$.
We are given $|\vec{d}\times\vec{c}|=14$, so $7|\lambda| = 14 \implies |\lambda| = 2$.
Finally, we need to find $|\vec{d}\cdot\vec{c}|$.
$\vec{d}\cdot\vec{c} = \lambda (\vec{a} \times \vec{b}) \cdot \vec{c}$. This is the scalar triple product.
$(\vec{a} \times \vec{b}) \cdot \vec{c} = (4\hat{i} + 3\hat{j} - \hat{k}) \cdot (6\hat{i}+3\hat{j}-2\hat{k}) = (4)(6) + (3)(3) + (-1)(-2) = 24+9+2 = 35$.
So, $\vec{d}\cdot\vec{c} = \lambda(35)$.
$|\vec{d}\cdot\vec{c}| = |\lambda(35)| = |\lambda| \cdot |35| = 2 \times 35 = 70$.