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Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVectors2 Marks
MCQ
If $\bar{a}, \bar{b}$ are unit vectors such that the vector $\overline{ a }+3 \overline{b}$ is perpendicular to $7 \overline{ a }-5 \overline{b}$ and $\overline{ a }-4 \overline{b}$ is perpendicular to $7 \bar{a}-2 \bar{b}$, then the angle between $\bar{a}$ and $\bar{b}$ is
  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • ✓
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$
✓

Answer

Correct option: C.
$\frac{\pi}{3}$
(C) Let $\theta$ be the angle between $\overline{ a }$ and $\overline{ b }$.
Now, $(\overline{ a }+3 \overline{b}) \perp(7 \overline{ a }-5 \overline{b})$
$\begin{array}{l}\Rightarrow(\overline{ a }+3 \overline{b}) \cdot(7 \overline{ a }-5 \overline{b})=0 \\ \Rightarrow 7|\overline{ a }|^2+16(\overline{ a } \cdot \overline{ b })-15|\overline{b}|^2=0 \\ \Rightarrow 7+16 \cos \theta-15=0 \\ \Rightarrow \cos \theta=\frac{1}{2} \\ \Rightarrow \theta=\frac{\pi}{3}\end{array}$
Also, $(\overline{ a }-4 \overline{b}) \perp(7 \overline{ a }-2 \overline{b})$
$\begin{array}{l}\Rightarrow(\overline{ a }-4 \overline{b}) \cdot(7 \overline{ a }-2 \overline{b})=0 \\ \Rightarrow 7|\overline{ a }|^2+8|b|^2-30(\overline{ a } \cdot \overline{ b })=0 \\ \Rightarrow 15-30 \cos \theta=0 \Rightarrow \cos \theta=\frac{1}{2} \\ \Rightarrow \theta=\frac{\pi}{3}\end{array}$

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