Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 17 maths textbook solution

Answers (1)

Answer:

 45°

Hint:

 cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}

Given:

 \frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \text { and } x+y+4=0

Solution:

x+y+4=0 \qquad \qquad \qquad \dots(i)

So the vector normal to the plane (i) is

\overrightarrow{n}=\overrightarrow{i}+\overrightarrow{j}

Cartesian equation of plane is

\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \qquad \qquad \qquad \dots(ii)

\overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}

Since

cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}

\begin{aligned} &cos(90-\theta )=\frac{2+1}{\sqrt{1+1}.\sqrt{4+1+4}}\\ &cos(90-\theta )=\frac{3}{\sqrt{2}.3}\\ &cos(90-\theta )=\frac{1}{\sqrt{2}}\\ &\Rightarrow sin\theta =\frac{1}{\sqrt{2}}\\ &\Rightarrow \theta =45^{\circ} \end{aligned}

 

Posted by

Gurleen Kaur

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 Class 12 Supplementary Result 2026: Scanned copy, verification, re-evaluation dates announced
anam-khan

SC asks CBSE to protect students affected by OSM evaluation in admissions
anam-khan

CBSE Class 12 Compartment Result 2026 LIVE: 12th supplementary marksheet out at results.digilocker.gov.in

Latest Question