First slide

Scalar triple product of vectors

Question

Let $\vec{a} \cdot \vec{b} = 0, where \vec{a} and \vec{b}$ are unit vectors and the unit vector $\vec{c}$ is inclined at an angle $θ$ to both $\vec{a} and \vec{b} . If \vec{c} = m \vec{a} + n \vec{b} + p (\vec{a} \times \vec{b}), (m, n, p \in R)$ then

Moderate

A

$- \frac{π}{4} \leq θ \leq \frac{π}{4}$
B

$\frac{π}{4} \leq θ \leq \frac{3 π}{4}$
C

$0 \leq θ \leq \frac{π}{4}$
D

$0 \leq θ \leq \frac{3 π}{4}$

Solution

$\vec{c} = m \vec{a} + \vec{nb} + p (\vec{a} \times \vec{b})$
Taking dot product with $\vec{a} and \vec{b}$ we have m=n=cos $θ$
$\Rightarrow | \vec{c} | = | \cos θ \vec{a} + \cos θ \vec{b} + p (\vec{a} \times \vec{b}) | = 1$
Squaring both sides, we get
$\begin{array}{l} \cos^{2} θ + \cos^{2} θ + p^{2} = 1 \\ or \cos θ = \pm \frac{\sqrt{1 - p^{2}}}{\sqrt{2}} \\ now - \frac{1}{\sqrt{2}} \leq \cos θ \leq \frac{1}{\sqrt{2}} (for real value of θ) \\ \frac{π}{4} \leq θ \leq \frac{3 π}{4} \end{array}$

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