Let a and b be two non-zero vectors and be the angle between then… - Vidyadip

MCQ

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors and $\theta$ be the angle between then.
Assertion (A) : $(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2 \neq|\vec{a}|^2|\vec{b}|^2$
Reason (R) : $\sin ^2 \theta+\cos ^2 \theta=1$

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
✓
(A) is false but (R) is true.

✓

Answer

Correct option: D.

(A) is false but (R) is true.

(d) :
$
\begin{aligned}
(\vec{a} \times \vec{b})^2 & +(\vec{a} \cdot \vec{b})^2=|\vec{a} \times \vec{b}|^2+(\vec{a} \cdot \vec{b})^2 \\
& =(|\vec{a}|| \vec{b} \mid \sin \theta)^2+(|\vec{a}||\vec{b}| \cos \theta)^2=|\vec{a}|^2|\vec{b}|^2
\end{aligned}
$
Hence, Assertion is false.
But $\sin ^2 \theta+\cos ^2 \theta=1$
Hence, reason is true.

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