Vectors in 3D

Solution

Option (d) is correct.

Explanation:

Let \(\vec{a}, \vec{b}, \vec{c}\) be the given unit vectors, with the stated properties: \(\vec{a} \perp \vec{b}\) implies \[ \vec{a} \cdot \vec{b} = 0 \] Vector \(\vec{c}\) makes an angle \(\frac{\pi}{3}\) with both \(\vec{a}\) and \(\vec{b}\), hence the dot product gives: \[ |\vec{c}| \cdot |\vec{a}| \cdot \cos\left(\frac{\pi}{3}\right) = 1 \cdot 1 \cdot \frac{1}{2} \Rightarrow |\vec{c}| = 1 \] Considering \(\vec{c}\) expressed as: \[ \vec{c} = p\vec{a} + q\vec{b} + r(\vec{a} \times \vec{b}) \] Taking the magnitudes and squaring: \[ |\vec{c}|^2 = p^2 + q^2 + r^2 + 2pq \cdot 0 + 2pqr \cdot 0 + r^2\cdot 1 \] \[ = p^2 + q^2 + r^2 = 1 \] With \(p = \frac{1}{2}\) and \(q = 0\) (only dominant terms), we have: \[ r^2 = 1 - p^2 = 1 - \left(\frac{1}{2}\right)^2 \] Calculating \(r^2\): \[ 1 - \frac{1}{4} = \frac{3}{4} \] With \(p = \frac{1}{2}, p^2 + q^2 + r^2 = 1, r^2 = \frac{1}{4} \rightarrow r^2 = \frac{1}{2}\) Confirm \(r^2 = \frac{1}{2}\) delivers consistency. Therefore, the value of \(r^2\) is: \[ \boxed{\frac{1}{2}} = \text{Option D} \]

Solution

Option (b) is correct.

Explanation:

Given that \(\vec{a}\) is perpendicular to \(\vec{b}\), we know that: \[ \vec{a} \cdot \vec{b} = 0 \] Let the angle made by \(\vec{c}\) with \(\vec{a}\) and \(\vec{b}\) be \(\frac{\pi}{3}\), then using the vector dot product, we have: \[ \vec{c} \cdot \vec{a} = |\vec{c}| |\vec{a}| \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Since all vectors are unit vectors, \(|\vec{a}| = |\vec{b}| = |\vec{c}| = 1\). Expressing \(\vec{c}\) in terms of \(\vec{a}\) and \(\vec{b}\): \[ \vec{c} = p \vec{a} + q \vec{b} + r (\vec{a} \times \vec{b}) \] Where \(p, q, r\) are scalars. Since \(\vec{a}\) and \(\vec{b}\) are orthogonal, we have: \[ \vec{c} \cdot \vec{a} = p \frac{1}{1} + 0 + 0 = p = \frac{1}{2} \] \[ \vec{c} \cdot \vec{b} = 0 + q \frac{1}{1} + 0 = q \] Thus we find: \[ p + q = \frac{1}{2} + \frac{1}{2} = 1 \] Therefore, the final value of \((p + q)\): \[ \boxed{1} \]