Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution for RD Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer Question 39

Provide Solution for RD Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer Question 39

Answers (1)

best_answer

Answer:  12

Given:  |\vec{a}|=5, \vec{a} \cdot \vec{b}=0 \text { and }|\vec{a}+\vec{b}|=13

Hint: Apply the identity  (a+b)^{2}

Solution:  |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \cdot \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}

\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]

\Rightarrow(13)^{2}=(5)^{2}+(|\vec{b}|)^{2}+2 \times 0

\left[\because|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=13,|\overrightarrow{\mathrm{a}}|^{2}=5 \& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0\right]

\begin{aligned} &\Rightarrow 169-25+|\vec{b}|^{2} \\ & \end{aligned}

\Rightarrow 169-25=|\vec{b}|^{2} \\

\Rightarrow|\vec{b}|^{2}=144 \\

\Rightarrow|\vec{b}|^{2}=12

 

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question