Question

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

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

Find the normal to the plane using cross product of \(\mathbf{e}_1 = (1, 0, 1)\) and \(\mathbf{e}_2 = (1, 1, 1)\): \[ \mathbf{n} = \mathbf{e}_1 \times \mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 0 & 1
1 & 1 & 1 \end{vmatrix} = (-1, 0, 1) \] Now project vector \(\mathbf{a} = (-1, 2, 0)\) on \(\mathbf{n}\): \[ \text{proj}_{\mathbf{n}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{n}}{|\mathbf{n}|^2} \cdot \mathbf{n} = \frac{(-1)(-1) + 0 + (0)(1)}{(-1)^2 + 1^2} \cdot (-1, 0, 1) = \frac{1}{2} \cdot (-1, 0, 1) = \left(-\frac{1}{2}, 0, \frac{1}{2}\right) \]