Question

Let $\vec{a} = \sqrt{7}\hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

• A. 44
• B. 54
• C. 86
• D. 132

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two vectors $\vec{a}$ and $\vec{b}$. We need to find a vector $\vec{r}$ that satisfies two conditions: one involving a cross product and another involving a dot product. Finally, we need to calculate the squared magnitude of the vector $3\vec{r}$.
Step 2: Key Formula or Approach:
1. Analyze the given vector equations to find an expression for $\vec{r}$.
2. The condition $\vec{u} \times \vec{v} = \vec{0}$ implies that vectors $\vec{u}$ and $\vec{v}$ are parallel, i.e., $\vec{u} = \lambda\vec{v}$ for some scalar $\lambda$.
3. The condition $\vec{r} \cdot \vec{a} = 0$ means that $\vec{r}$ is perpendicular to $\vec{a}$.
4. We will use these properties to find $\vec{r}$ in terms of $\vec{a}$ and $\vec{b}$.
Step 3: Detailed Explanation:
We are given two conditions for the vector $\vec{r}$:
1) $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$
2) $\vec{r} \cdot \vec{a} = 0$
Let's work with the first condition. Using the anti-commutative property of the cross product ($\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$), we have $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.
The equation becomes: $\vec{r} \times \vec{a} - (\vec{b} \times \vec{a}) = \vec{0}$.
Using the distributive property of the cross product:
$(\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$.
This implies that the vector $(\vec{r} - \vec{b})$ is parallel to the vector $\vec{a}$. Therefore, we can write:
$\vec{r} - \vec{b} = \lambda \vec{a}$ for some scalar $\lambda$.
This gives us an expression for $\vec{r}$:
$\vec{r} = \vec{b} + \lambda \vec{a}$.
Now, we use the second condition, $\vec{r} \cdot \vec{a} = 0$.
Substitute the expression for $\vec{r}$:
$(\vec{b} + \lambda \vec{a}) \cdot \vec{a} = 0$
$\vec{b} \cdot \vec{a} + \lambda(\vec{a} \cdot \vec{a}) = 0$
$\vec{b} \cdot \vec{a} + \lambda|\vec{a}|^2 = 0$
Solving for $\lambda$:
$\lambda = -\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}$.
Let's calculate the required dot product and squared magnitude from the given vectors:
$\vec{a} = \sqrt{7}\hat{i}+\hat{j}-\hat{k}$
$\vec{b} = 0\hat{i}+\hat{j}+2\hat{k}$
$\vec{b} \cdot \vec{a} = (0)(\sqrt{7}) + (1)(1) + (2)(-1) = 0 + 1 - 2 = -1$.
$|\vec{a}|^2 = (\sqrt{7})^2 + 1^2 + (-1)^2 = 7 + 1 + 1 = 9$.
Substituting these values to find $\lambda$:
$\lambda = -\frac{-1}{9} = \frac{1}{9}$.
Now we have $\vec{r} = \vec{b} + \frac{1}{9}\vec{a}$.
We need to find $|3\vec{r}|^2$.
$|3\vec{r}|^2 = (3\vec{r}) \cdot (3\vec{r}) = 9(\vec{r} \cdot \vec{r}) = 9|\vec{r}|^2$.
Let's find $|\vec{r}|^2$:
$|\vec{r}|^2 = \vec{r} \cdot \vec{r} = (\vec{b} + \lambda \vec{a}) \cdot (\vec{b} + \lambda \vec{a})$
$= |\vec{b}|^2 + 2\lambda(\vec{a} \cdot \vec{b}) + \lambda^2|\vec{a}|^2$.
We need $|\vec{b}|^2$:
$|\vec{b}|^2 = 0^2 + 1^2 + 2^2 = 1 + 4 = 5$.
Substitute the values of $|\vec{b}|^2$, $\vec{a} \cdot \vec{b}$, $\lambda$, and $|\vec{a}|^2$:
$|\vec{r}|^2 = 5 + 2\left(\frac{1}{9}\right)(-1) + \left(\frac{1}{9}\right)^2(9)$
$|\vec{r}|^2 = 5 - \frac{2}{9} + \frac{1}{81} \times 9 = 5 - \frac{2}{9} + \frac{1}{9} = 5 - \frac{1}{9} = \frac{44}{9}$.
Finally, calculate $|3\vec{r}|^2$:
$|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \times \frac{44}{9} = 44$.
Step 4: Final Answer:
The value of $|3\vec{r}|^2$ is 44.