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 \(2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = 0\). If \(\vec{a} \cdot \vec{c} = 15\), then the value of \(\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})\) is

• A. 5
• B. -5
• C. 3
• D. -3

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use properties of the cross product to find $\vec{c}$ in terms of $\vec{a}$ and $\vec{b}$, then use the dot product condition $\vec{a}\cdot\vec{c}=15$ to find a scalar $\lambda$, and finally compute $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$.
Step 2: Extracting the relation for $\vec{c$:}
Given: \[ 2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0} \] \[ 2(\vec{a}\times\vec{b}) - 3(\vec{c}\times\vec{b}) = \vec{0} \] \[ (2\vec{a} - 3\vec{c})\times\vec{b} = \vec{0} \] This means $\vec{b} \parallel (2\vec{a} - 3\vec{c})$, so: \[ 2\vec{a} - 3\vec{c} = \lambda\vec{b} \] for some scalar $\lambda$.
Therefore: \[ \vec{c} = \frac{2\vec{a} - \lambda\vec{b}}{3} \tag{1} \] Step 3: Using the dot product condition to find $\lambda$:
\[ \vec{a}\cdot\vec{c} = 15 \implies \vec{a}\cdot\frac{2\vec{a}-\lambda\vec{b}}{3} = 15 \] \[ \frac{2|\vec{a}|^2 - \lambda(\vec{a}\cdot\vec{b})}{3} = 15 \] Now compute $|\vec{a}|^2$ and $\vec{a}\cdot\vec{b}$: \[ |\vec{a}|^2 = 4^2 + (-1)^2 + 3^2 = 16 + 1 + 9 = 26 \] \[ \vec{a}\cdot\vec{b} = (4)(10) + (-1)(2) + (3)(-1) = 40 - 2 - 3 = 35 \] Substituting: \[ \frac{2(26) - 35\lambda}{3} = 15 \implies 52 - 35\lambda = 45 \implies 35\lambda = 7 \implies \lambda = \frac{1}{5} \] Step 4: Computing $\vec{c\cdot(\hat{i}+\hat{j}-3\hat{k})$:}
From equation (1): \[ \vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k}) = \frac{(2\vec{a} - \lambda\vec{b})\cdot(\hat{i}+\hat{j}-3\hat{k})}{3} \] Compute $\vec{a}\cdot(\hat{i}+\hat{j}-3\hat{k})$: \[ = 4(1) + (-1)(1) + 3(-3) = 4 - 1 - 9 = -6 \] Compute $\vec{b}\cdot(\hat{i}+\hat{j}-3\hat{k})$: \[ = 10(1) + 2(1) + (-1)(-3) = 10 + 2 + 3 = 15 \] Therefore: \[ \vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k}) = \frac{2(-6) - \frac{1}{5}(15)}{3} = \frac{-12 - 3}{3} = \frac{-15}{3} = -5 \] Step 5: Final Answer:
\[ \vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k}) = -5 \] The answer is Option (2).