Short Quiz - Chapter 10 Class 12 Vector Algebra

Correct option: A

Answer: For any two vectors:

$$ \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta $$


Since:

$$ \cos \theta \leq 1 $$


Therefore:

$$ \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}| $$


This is always true.
Answer: A) \(\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|\)

Correct option: C

Answer: $$ \vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \quad \vec{b}=\hat{i}+\hat{j}-\hat{k} $$


Check dot product:

$$ \begin{gathered} \vec{a} \cdot \vec{b}=(2)(1)+(-1)(1)+(1)(-1) \\ =2-1-1=0 \end{gathered} $$


Since dot product is \(\mathbf{0}\), vectors are perpendicular.
Answer: C) perpendicular vectors

Correct option: D

Answer: Given:

$$ |\vec{a}|=2, \quad-3 \leq k \leq 2 $$


We need:

$$ |k \vec{a}| $$


Formula:

$$ \begin{gathered} |k \vec{a}|=|k||\vec{a}| \\ =2|k| \end{gathered} $$


Since:

$$ \begin{gathered} -3 \leq k \leq 2 \\ |k| \in[0,3] \end{gathered} $$


So:

$$ 2|k| \in[0,6] $$


Answer: D) [0, 6]

Correct option: A

Answer: Assertion (A): The vectors \(a\) and ( \(-2 a\) ), where \(a \neq 0\), are collinear vectors.
Reason (R): \(a \cdot(-2 a)=0\).
Step-by-Step Reasoning:
1. Analyze Assertion (A): Two vectors are collinear if one can be expressed as a scalar multiple of the other ( \(u=k v\) ). Since ( \(-2 a\) ) is simply -2 times the vector \(a\), they are by definition collinear (they lie on the same or parallel lines). Therefore, Assertion (A) is True.
2. Analyze Reason (R): Let's calculate the dot product \(a \cdot(-2 a)\) :

$$ a \cdot(-2 a)=-2(a \cdot a)=-2|a|^2 $$


Since we are given that \(a \neq 0,|a|^2\) must be greater than zero. Therefore, the dot product is \(-2|a|^2\), which is not equal to 0 . The only way a dot product is zero is if the vectors are perpendicular (orthogonal), which these are not. Therefore, Reason (R) is False.

Answer: C) A is true, but R is false.