Question

Let \(\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}\), \(\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}\) and a vector \(\vec{c}\) be such that \(2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0}\). If \(\vec{a} \cdot \vec{c} = 15\), then \(\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})\) is equal to:

• A. -6
• B. -5
• C. -4
• D. -3

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given a relationship involving cross products of three vectors. By using the anti-commutative property of cross products, we can express one vector as a linear combination of the other two. We use the dot product given to find the scalar constant in the combination.
Step 2: Key Formula or Approach:
Cross product properties: \( \vec{b} \times \vec{c} = -(\vec{c} \times \vec{b}) \).
If \( \vec{u} \times \vec{v} = \vec{0} \), then vectors \( \vec{u} \text{ and } \vec{v} \) are collinear, meaning \( \vec{u} = \lambda \vec{v} \).
Step 3: Detailed Explanation:
Given equation: \( 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0} \).
Use property of cross product to reverse the second term:
\( 2(\vec{a} \times \vec{b}) - 3(\vec{c} \times \vec{b}) = \vec{0} \).
Factor out \( \times \vec{b} \) on the right side:
\( (2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0} \).
Since the cross product of two vectors is zero, they must be parallel (or one is the zero vector). Assuming \( \vec{b} \neq 0 \):
\( 2\vec{a} - 3\vec{c} = \lambda \vec{b} \) for some scalar \( \lambda \).
Rearrange to solve for \( \vec{c} \):
\( 3\vec{c} = 2\vec{a} - \lambda \vec{b} \implies \vec{c} = \frac{2}{3}\vec{a} - \frac{\lambda}{3}\vec{b} \).
We are given the dot product condition: \( \vec{a} \cdot \vec{c} = 15 \).
Take the dot product of both sides with \( \vec{a} \):
\( \vec{a} \cdot \vec{c} = \frac{2}{3} (\vec{a} \cdot \vec{a}) - \frac{\lambda}{3} (\vec{a} \cdot \vec{b}) \).
Let's compute the necessary dot products from the given vectors:
\( \vec{a} = 4\hat{i} - \hat{j} + 3\hat{k} \)
\( \vec{b} = 10\hat{i} + 2\hat{j} - \hat{k} \)
\( |\vec{a}|^2 = \vec{a} \cdot \vec{a} = (4)^2 + (-1)^2 + (3)^2 = 16 + 1 + 9 = 26 \).
\( \vec{a} \cdot \vec{b} = (4)(10) + (-1)(2) + (3)(-1) = 40 - 2 - 3 = 35 \).
Substitute these into our equation:
\( 15 = \frac{2}{3} (26) - \frac{\lambda}{3} (35) \).
Multiply the entire equation by 3 to remove fractions:
\( 45 = 52 - 35\lambda \).
\( 35\lambda = 52 - 45 = 7 \implies \lambda = \frac{7}{35} = \frac{1}{5} \).
Now we can write the explicit form of \( \vec{c} \):
\( \vec{c} = \frac{2}{3}\vec{a} - \frac{1/5}{3}\vec{b} = \frac{2}{3}\vec{a} - \frac{1}{15}\vec{b} \).
We need to evaluate \( \vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k}) \).
Let \( \vec{v} = \hat{i} + \hat{j} - 3\hat{k} \).
We want to find \( \vec{c} \cdot \vec{v} = \left( \frac{2}{3}\vec{a} - \frac{1}{15}\vec{b} \right) \cdot \vec{v} = \frac{2}{3}(\vec{a} \cdot \vec{v}) - \frac{1}{15}(\vec{b} \cdot \vec{v}) \).
Calculate the individual dot products:
\( \vec{a} \cdot \vec{v} = (4)(1) + (-1)(1) + (3)(-3) = 4 - 1 - 9 = -6 \).
\( \vec{b} \cdot \vec{v} = (10)(1) + (2)(1) + (-1)(-3) = 10 + 2 + 3 = 15 \).
Substitute these back:
\( \vec{c} \cdot \vec{v} = \frac{2}{3}(-6) - \frac{1}{15}(15) \).
\( = 2(-2) - 1 = -4 - 1 = -5 \).
Step 4: Final Answer:
The value is -5.