First slide

Scalar triple product of vectors

Question

Let $\vec{a}, \vec{b} and \vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector, which is perpendicular to $(\vec{a} + \vec{b} + \vec{c}) . Now \vec{d} = (\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + 2 (\vec{c} \times \vec{a})$ . then

Moderate

A

$\frac{\vec{d} \cdot (\vec{a} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = 2$
B

$\frac{\vec{d} \cdot (\vec{a} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = - 2$
C

minimum value of $x^{2} + y^{2} is π^{2} / 4$
D

minimum value of $x^{2} + y^{2} is 5 π^{2} / 4$

Solution

$\begin{array}{l} \vec{d} \cdot \vec{a} = [\vec{a} \vec{b} \vec{c}] \cos y = - \vec{d} \cdot (\vec{b} + \vec{c}) \\ or \cos y = - \frac{\vec{d} \cdot (\vec{b} + \vec{c})}{[\vec{a,} \vec{b}, \vec{c}]} \\ Similarly, \sin x = \frac{\vec{d} \cdot (\vec{a} + \vec{b})}{[\vec{a} \vec{b} \vec{c}]} and \frac{\vec{d} \cdot (\vec{a} + \vec{c})}{[\vec{a} \vec{b} \vec{c}]} = - 2 \\ ∴ \sin x + \cos y + 2 = 0 \\ or \sin x + \cos y = - 2 \\ or \sin x = - 1, \cos y = - 1 \end{array}$
Since we want the minimum value of x²+ y² , $x = - π / 2, y = π$
minimum value of $x^{2} + y^{2} is 5 π^{2} / 4$

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