Find the value of the constant k for which \(a = 4 i - k j\) and \(b = 3 i + 8 j\) are perpendicular.

Answer Details

For two vectors to be perpendicular, their dot product must be zero. So, we can find the value of \(k\) by taking the dot product of the given vectors and setting it equal to zero. The dot product of two vectors, \(\vec{a} \cdot \vec{b}\), is given by: \(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z\) where \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) and \(\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}\) are the components of the vectors along the x, y, and z axes. So, using the given vectors, we have: \(\vec{a} \cdot \vec{b} = (4\hat{i} - k\hat{j}) \cdot (3\hat{i} + 8\hat{j}) = 4 \times 3 + (-k) \times 8 = 12 - 8k\) For the vectors to be perpendicular, this dot product must be zero, so: \(12 - 8k = 0\) Solving for \(k\), we get: \(k = \frac{12}{8} = \frac{3}{2}\) Therefore, the answer is option (D) \(\frac{3}{2}\). So, the constant \(k\) must be equal to \(\frac{3}{2}\) for the vectors \(a\) and \(b\) to be perpendicular.