Question

Let a =\(\^i+4\^j​+2\^k,b=3\^i−2\^j​+7\^k \,\,\,\,and\,\,\.\ c=2\^i−\^j​+4\^k\). Find a vector d satisfies \(\overrightarrow{d}\times{\overrightarrow{b}}=\overrightarrow{c}\times{\overrightarrow{b}}\) and \(\overrightarrow{d}.\overrightarrow{a}=24\), then \(|\overrightarrow{d}|^2\) is equal to

• A. 323
• B. 313
• C. 423
• D. 413

Answer 1

The Correct Option is D

Solution:
From \(\mathbf{d} \times \mathbf{b} = \mathbf{c} \times \mathbf{b}\), we get \[ (\mathbf{d} - \mathbf{c}) \times \mathbf{b} = \mathbf{0} \quad \Longrightarrow \quad \mathbf{d} = \mathbf{c} + \lambda \mathbf{b}. \] Since \(\mathbf{d} \cdot \mathbf{a} = 24\), we substitute \(\mathbf{d} = \mathbf{c} + \lambda \mathbf{b}\) to obtain \[ (\mathbf{c} + \lambda \mathbf{b}) \cdot \mathbf{a} = 24. \] Calculate \(\mathbf{a} \cdot \mathbf{c}\) and \(\mathbf{a} \cdot \mathbf{b}\): \[ \mathbf{a} \cdot \mathbf{c} = (1)(2) + (4)(-1) + (2)(4) = 2 - 4 + 8 = 6, \] \[ \mathbf{a} \cdot \mathbf{b} = (1)(3) + (4)(-2) + (2)(7) = 3 - 8 + 14 = 9. \] Hence \[ (\mathbf{c} + \lambda \mathbf{b}) \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{a} + \lambda (\mathbf{b} \cdot \mathbf{a}) = 6 + 9\lambda = 24 \quad \Longrightarrow \quad \lambda = 2. \] Therefore \[ \mathbf{d} = \mathbf{c} + 2 \mathbf{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + 2(3\hat{i} - 2\hat{j} + 7\hat{k}) = 8\hat{i} - 5\hat{j} + 18\hat{k}. \] Finally, \[ |\mathbf{d}|^2 = 8^2 + (-5)^2 + 18^2 = 64 + 25 + 324 = 413. \]