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 looking for a vector $\vec{r}$ that satisfies two conditions: a cross product relation and a dot product relation with given vectors $\vec{a}$ and $\vec{b}$. After finding $\vec{r}$, we must compute the squared magnitude of $3\vec{r}$.
Step 2: Key Formula or Approach:
1. Analyze the Cross Product Equation: Rearrange the first equation using vector properties, specifically $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ and the distributive law, to simplify it.
2. Interpret Geometrically: The condition $\vec{X} \times \vec{Y} = \vec{0}$ means $\vec{X}$ is parallel to $\vec{Y}$ ($\vec{X} = \lambda \vec{Y}$). The condition $\vec{X} \cdot \vec{Y} = 0$ means they are perpendicular.
3. Solve for $\vec{r$:} Use the simplified cross product relation to express $\vec{r}$ in terms of $\vec{a}$, $\vec{b}$ and a scalar $\lambda$. Then use the dot product condition to solve for $\lambda$.
4. Calculate the Result: Substitute the value of $\lambda$ back to define $\vec{r}$ and then calculate $|3\vec{r}|^2 = 9|\vec{r}|^2$.
Step 3: Detailed Explanation:
We are given the two equations:
1) $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$
2) $\vec{r} \cdot \vec{a} = 0$
Let's simplify the first equation. We can rewrite it as $\vec{r} \times \vec{a} = -(\vec{a} \times \vec{b})$. Using the anti-commutative property, this is $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$.
Rearranging this gives:
\[ \vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0} \] \[ (\vec{r} - \vec{b}) \times \vec{a} = \vec{0} \] This equation implies that the vector $(\vec{r} - \vec{b})$ is parallel to vector $\vec{a}$. Therefore, we can write:
\[ \vec{r} - \vec{b} = \lambda \vec{a} \] for some scalar $\lambda$. This gives an expression for $\vec{r}$:
\[ \vec{r} = \vec{b} + \lambda \vec{a} \] Now, apply the second condition, $\vec{r} \cdot \vec{a} = 0$:
\[ (\vec{b} + \lambda \vec{a}) \cdot \vec{a} = 0 \] \[ \vec{b} \cdot \vec{a} + \lambda (\vec{a} \cdot \vec{a}) = 0 \] \[ \vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2 = 0 \] We can now solve for $\lambda$:
\[ \lambda = - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \] Let's calculate the necessary scalar values from the given vectors:
$\vec{a} = \sqrt{7}\hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$.
\[ \vec{b} \cdot \vec{a} = (\sqrt{7})(0) + (1)(1) + (-1)(2) = -1 \] \[ |\vec{a}|^2 = (\sqrt{7})^2 + 1^2 + (-1)^2 = 7 + 1 + 1 = 9 \] Substitute these to find $\lambda$:
\[ \lambda = - \frac{-1}{9} = \frac{1}{9} \] Now we have a complete expression for $\vec{r}$: $\vec{r} = \vec{b} + \frac{1}{9}\vec{a}$. We need to find $|3\vec{r}|^2 = 9|\vec{r}|^2$. Let's compute $|\vec{r}|^2$:
\[ |\vec{r}|^2 = \vec{r} \cdot \vec{r} = (\vec{b} + \lambda\vec{a}) \cdot (\vec{b} + \lambda\vec{a}) = |\vec{b}|^2 + 2\lambda(\vec{a}\cdot\vec{b}) + \lambda^2|\vec{a}|^2 \] We need $|\vec{b}|^2 = 0^2 + 1^2 + 2^2 = 5$.
\[ |\vec{r}|^2 = 5 + 2\left(\frac{1}{9}\right)(-1) + \left(\frac{1}{9}\right)^2(9) = 5 - \frac{2}{9} + \frac{1}{9} = 5 - \frac{1}{9} = \frac{44}{9} \] Finally, the required value is:
\[ |3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \left(\frac{44}{9}\right) = 44 \] Step 4: Final Answer:
The value of $|3\vec{r}|^2$ is 44.