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Please solve RD Sharma class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 21 maths textbook solution

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Answer:- k \in\left(-1, \frac{1}{2}\right) \cup\left(-\frac{1}{2}, 1\right)  

Hint:-To solve this question, we use magnitude formula.

Given:-The value of k for which |k \vec{a}|<|\vec{a}| \text { and } k \vec{a}+\frac{1}{2} \vec{a}is parallel to \vec{a} holds true are ---

Solution:- |k \vec{a}|<|\vec{a}|

\begin{aligned} &\Rightarrow k|\vec{a}|<|\vec{a}| \\\\ &\Rightarrow k<1 \\ \\&\Rightarrow-1<k<1 k \in(-1,1) \end{aligned}\begin{aligned} &\Rightarrow k|\vec{a}|<|\vec{a}| \\\\ &\Rightarrow k<1 \\ \\&\Rightarrow-1<k<1 k \in(-1,1) \end{aligned}

\begin{aligned} &k \vec{a}+\frac{1}{2} \vec{a} \text { is parallel to } a\\\\ &\text { If we put } k=\frac{-1}{2} \text { then } k \vec{a}+\frac{1}{2} \vec{a} \end{aligned}

\begin{aligned} &=-\frac{1}{2} \vec{a}+\frac{1}{2} \vec{a}=0 \\\\ &k \neq \frac{-1}{2} \end{aligned}

k \in\left(-1, \frac{1}{2}\right) \cup\left(-\frac{1}{2}, 1\right)

 

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