Question

Find the component of the vector \((-1, 2, 0)\) perpendicular to the plane of the vectors \(\mathbf{e}_1 = (1, 0, 1)\) and \(\mathbf{e}_2 = (1, 1, 1)\).

• A. \(\left( -\frac{1}{2}, \frac{1}{2}, 0 \right)\)
• B. \(\left( 0, -\frac{1}{2}, \frac{1}{2} \right)\)
• C. \(\left( -\frac{1}{2}, 0, \frac{1}{2} \right)\)
• D. \(\left( \frac{1}{2}, 0, -\frac{1}{2} \right)\)

Answer 1

The Correct Option is C

The direction vector perpendicular to both \(\mathbf{e}_1\) and \(\mathbf{e}_2\) is their cross product: \[ \mathbf{n} = \mathbf{e}_1 \times \mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 0 & 1
1 & 1 & 1 \end{vmatrix} = (0 - 1)\mathbf{i} - (1 - 1)\mathbf{j} + (1 - 0)\mathbf{k} = -\mathbf{i} + \mathbf{k} \Rightarrow \mathbf{n} = (-1, 0, 1) \] Let projection of \(\mathbf{a}\) on \(\mathbf{n}\): \[ \text{proj}_{\mathbf{n}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{n}}{|\mathbf{n}|^2} \cdot \mathbf{n} = \frac{(-1,2,0) \cdot (-1,0,1)}{(-1)^2 + 0^2 + 1^2} \cdot (-1,0,1) = \frac{1}{2} \cdot (-1,0,1) = \left(-\frac{1}{2}, 0, \frac{1}{2}\right) \]