Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a} + \vec{b}| = 15$ and
\( \vec{a} \times (3\hat{i} - 4\hat{j} + 5\hat{k}) = (3\hat{i} - 4\hat{j} + 5\hat{k}) \times \vec{b} \)
What is the value of $|(\vec{a} + \vec{b}) \cdot (2\hat{i} + 3\hat{j} + \hat{k})|$?
Show Hint
Whenever you see $\vec{u} \times \vec{w} = \vec{w} \times \vec{v}$, immediately rewrite it as $(\vec{u} + \vec{v}) \times \vec{w} = \vec{0}$.
This geometric identity instantly shows that the sum vector $(\vec{u} + \vec{v})$ is parallel to $\vec{w}$.
Step 1 : Understanding the Question:
The question asks us to compute the absolute value of the dot product of the vector sum $(\vec{a} + \vec{b})$ with the vector $(2\hat{i} + 3\hat{j} + \hat{k})$, based on a given cross-product relation. Step 2 : Key Formulas and Approach:
Let $\vec{c} = 3\hat{i} - 4\hat{j} + 5\hat{k}$.
The given equation is $\vec{a} \times \vec{c} = \vec{c} \times \vec{b}$.
Using the anticommutative property of the cross product ($\vec{c} \times \vec{b} = -\vec{b} \times \vec{c}$), we will rewrite this as a single cross-product equation.
This will show that $(\vec{a} + \vec{b})$ is collinear (parallel) to $\vec{c}$, allowing us to express it as $\lambda \vec{c}$. Step 3 : Detailed Explanation:
Let us simplify the cross-product relation:
\[ \vec{a} \times \vec{c} = -\vec{b} \times \vec{c} \]
\[ \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0} \]
\[ (\vec{a} + \vec{b}) \times \vec{c} = \vec{0} \]
Since the cross product of $(\vec{a} + \vec{b})$ and $\vec{c}$ is the zero vector, they must be parallel:
\[ \vec{a} + \vec{b} = \lambda \vec{c} \]
where $\lambda$ is a real scalar.
Let us find $|\lambda|$ using the given modulus $|\vec{a} + \vec{b}| = 15$:
\[ |\vec{a} + \vec{b}| = |\lambda \vec{c}| = |\lambda| |\vec{c}| \]
First, calculate $|\vec{c}|$:
\[ |\vec{c}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]
Now, substitute this into the modulus equation:
\[ 15 = |\lambda| \cdot 5\sqrt{2} \implies |\lambda| = \frac{15}{5\sqrt{2}} = \frac{3}{\sqrt{2}} \]
Let $\vec{d} = 2\hat{i} + 3\hat{j} + \hat{k}$. We want to evaluate:
\[ |(\vec{a} + \vec{b}) \cdot \vec{d}| = |\lambda \vec{c} \cdot \vec{d}| = |\lambda| |\vec{c} \cdot \vec{d}| \]
Let us calculate the dot product $\vec{c} \cdot \vec{d}$:
\[ \vec{c} \cdot \vec{d} = (3\hat{i} - 4\hat{j} + 5\hat{k}) \cdot (2\hat{i} + 3\hat{j} + \hat{k}) \]
\[ \vec{c} \cdot \vec{d} = 3(2) + (-4)(3) + 5(1) \]
\[ \vec{c} \cdot \vec{d} = 6 - 12 + 5 = -1 \]
Now substitute the values back:
\[ |(\vec{a} + \vec{b}) \cdot \vec{d}| = \frac{3}{\sqrt{2}} \times |-1| = \frac{3}{\sqrt{2}} \] Step 4 : Final Answer:
The value of the expression is $\frac{3}{\sqrt{2}}$.
This corresponds to Option (A).
