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 \( |\vec{a} \cdot \vec{d}| \).
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 \( |\vec{a} \cdot \vec{d}| \).
Show Hint
- 18
- 12
- 9
- 15
The Correct Option is D
Approach Solution - 1
Approach Solution -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
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