Question

Let \( \vec{a} = 2\hat{i} - 3\hat{j} + \hat{k} \), \( \vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k} \) and a vector \( \vec{c} \) be such that \[ (\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k} \] and \[ \vec{a} \cdot \vec{c} = 3. \] If \( \vec{b} \times \vec{c} = \vec{d} \), then find \( |\vec{a} \cdot \vec{d}| \).

• A. 18
• B. 12
• C. 9
• D. 15

Hint:

When working with cross and dot products, make sure to calculate each component carefully and use the appropriate properties of these operations to find relationships between vectors.

Answer 1

The Correct Option is D

Given: \[ \mathbf{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \mathbf{b} = 3\hat{i} + 2\hat{j} + 5\hat{k} \] The cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as follows: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & -3 & 1 3 & 2 & 5 \end{vmatrix} = \hat{i} \left( (-3)(5) - (1)(2) \right) - \hat{j} \left( (2)(5) - (1)(3) \right) + \hat{k} \left( (2)(2) - (-3)(3) \right) \] \[ = \hat{i}(-15 - 2) - \hat{j}(10 - 3) + \hat{k}(4 + 9) \] \[ = -17\hat{i} - 7\hat{j} + 13\hat{k} \] So, \[ \mathbf{a} \times \mathbf{b} = -17\hat{i} - 7\hat{j} + 13\hat{k} \] We are given that: \[ (\mathbf{a} - \mathbf{c}) \times \mathbf{b} = -18\hat{i} - 3\hat{j} + 12\hat{k} \] Thus, we have: \[ (\mathbf{a} - \mathbf{c}) \times \mathbf{b} = \mathbf{a} \times \mathbf{b} - \mathbf{c} \times \mathbf{b} \] \[ -18\hat{i} - 3\hat{j} + 12\hat{k} = -17\hat{i} - 7\hat{j} + 13\hat{k} - \mathbf{c} \times \mathbf{b} \] So, \[ \mathbf{c} \times \mathbf{b} = -\hat{i} + 4\hat{j} - \hat{k} \] Now, we use the condition \( \mathbf{b} \times \mathbf{c} = \mathbf{a} \): \[ \mathbf{b} \times \mathbf{c} = (3\hat{i} + 2\hat{j} + 5\hat{k}) \times \mathbf{c} = 2\hat{i} - 3\hat{j} + \hat{k} \] Thus, \[ \mathbf{c} \times \mathbf{b} = \mathbf{a} \implies \mathbf{c} = \left( \mathbf{a} \cdot \mathbf{b} \right) \] Finally, we calculate: \[ \mathbf{a} \cdot \mathbf{c} = -2 - 12 - 1 = -15 \] Hence, \( |\mathbf{a} \cdot \mathbf{c}| = 15 \).

Answer 2

We have vectors \(\mathbf{a}=\langle 2,-3,1\rangle\) and \(\mathbf{b}=\langle 3,2,5\rangle\). A vector \(\mathbf{c}\) satisfies \((\mathbf{a}-\mathbf{c})\times\mathbf{b}=\langle -18,-3,12\rangle\) and \(\mathbf{a}\cdot\mathbf{c}=3\). Define \(\mathbf{d}=\mathbf{b}\times\mathbf{c}\). We need \( |\mathbf{a}\cdot\mathbf{d}| \).

Concept Used:

Use the identity \((\mathbf{a}-\mathbf{c})\times\mathbf{b}=\mathbf{a}\times\mathbf{b}-\mathbf{c}\times\mathbf{b}=\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}=\mathbf{a}\times\mathbf{b}+\mathbf{d}\).

Step-by-Step Solution:

Step 1: Compute \(\mathbf{a}\times\mathbf{b}\).

\[ \mathbf{a}\times\mathbf{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 2&-3&1\\ 3&2&5 \end{vmatrix} = (-17)\hat{i}+(-7)\hat{j}+13\hat{k}=\langle -17,-7,13\rangle. \]

Step 2: Use \((\mathbf{a}-\mathbf{c})\times\mathbf{b}=\mathbf{a}\times\mathbf{b}+\mathbf{d}\) to find \(\mathbf{d}\).

\[ \mathbf{d}=\langle -18,-3,12\rangle-\langle -17,-7,13\rangle=\langle -1,4,-1\rangle. \]

Step 3: Compute \(\mathbf{a}\cdot\mathbf{d}\).

\[ \mathbf{a}\cdot\mathbf{d}=\langle 2,-3,1\rangle\cdot\langle -1,4,-1\rangle =2(-1)+(-3)(4)+1(-1)=-2-12-1=-15. \]

Final Computation & Result

\[ |\mathbf{a}\cdot\mathbf{d}|=|-15|=15. \]

Answer: 15