Question

The vectors are $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}, \vec{b} = \hat{i} + \hat{j}$ if $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$ and $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $\frac{\pi}{4}$ then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is

• A. 3
• B. $\frac{3}{\sqrt{2}}$
• C. $3\sqrt{2}$
• D. 1

Answer 1

Answer: (B)

$\frac{3}{\sqrt{2}}$

Answer 2

To find $|(\vec{a} \times \vec{b}) \times \vec{c}|$, we need to calculate the magnitudes of $\vec{a} \times \vec{b}$ and $\vec{c}$, and the sine of the angle between them.

Step 1: Calculate $\vec{a} \times \vec{b}$ and its magnitude. Given $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. The cross product $\vec{a} \times \vec{b}$ is:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$ $$= \hat{i}((1)(0) - (-2)(1)) - \hat{j}((2)(0) - (-2)(1)) + \hat{k}((2)(1) - (1)(1))$$ $$= \hat{i}(0 + 2) - \hat{j}(0 + 2) + \hat{k}(2 - 1)$$ $$= 2\hat{i} - 2\hat{j} + \hat{k}$$

Let $\vec{d} = \vec{a} \times \vec{b}$. The magnitude of $\vec{d}$ is:

$$|\vec{d}| = |2\hat{i} - 2\hat{j} + \hat{k}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

Step 2: Find the magnitude of $\vec{c}$ using the given conditions. We are given two conditions for $\vec{c}$:

1. $\vec{a} \cdot \vec{c} = |\vec{c}|$
2. $|\vec{c} - \vec{a}| = 2\sqrt{2}$

First, let's find the magnitude of $\vec{a}$:

$$|\vec{a}| = |2\hat{i} + \hat{j} - 2\hat{k}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Now, square the second condition:

$$|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2$$

Using the property $|\vec{x}|^2 = \vec{x} \cdot \vec{x}$:

$$(\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8$$ $$|\vec{c}|^2 - 2(\vec{c} \cdot \vec{a}) + |\vec{a}|^2 = 8$$

Substitute the first condition ($\vec{c} \cdot \vec{a} = |\vec{c}|$) and the value of $|\vec{a}| = 3$:

$$|\vec{c}|^2 - 2|\vec{c}| + 3^2 = 8$$ $$|\vec{c}|^2 - 2|\vec{c}| + 9 = 8$$

Rearrange the terms to form a quadratic equation in $|\vec{c}|$:

$$|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$$

This is a perfect square trinomial:

$$(|\vec{c}| - 1)^2 = 0$$

Therefore,

$$|\vec{c}| = 1$$

Step 3: Calculate $|(\vec{a} \times \vec{b}) \times \vec{c}|$. We need to find $|\vec{d} \times \vec{c}|$. The magnitude of the cross product of two vectors is given by $|\vec{X} \times \vec{Y}| = |\vec{X}| |\vec{Y}| \sin \theta$, where $\theta$ is the angle between $\vec{X}$ and $\vec{Y}$. Here, $\vec{X} = \vec{d} = \vec{a} \times \vec{b}$ and $\vec{Y} = \vec{c}$. We found $|\vec{d}| = 3$ and $|\vec{c}| = 1$. The angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is given as $\frac{\pi}{4}$. So, $\theta = \frac{\pi}{4}$.

$$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{d}| |\vec{c}| \sin\left(\frac{\pi}{4}\right)$$ $$= (3)(1)\left(\frac{1}{\sqrt{2}}\right)$$ $$= \frac{3}{\sqrt{2}}$$