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 R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 3 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 3 Maths Textbook Solution.

Answers (1)

Answer:Therefore ,the distance is 13 Units .

Hint : Use Distance formula \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-_y{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

Given:

\begin{aligned} &\vec{r}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k}) \text { and } \\ &\vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=5 \text { and point }(-1,-5,-10) \end{aligned}

Solution :

General\; point \: on \: the \: line \: is

(2+3 \lambda,-1+4 \lambda, 2+2 \lambda)

As\: point \: lies \: on\: the\: plane

\therefore 2+3 \lambda+1-4 \lambda+2+2 \lambda=5

\Rightarrow \lambda=0

    \therefore \: So\: the \: \: coordinate \: \: of \: \: point \: \: is (2,-1,2)

Distance =\sqrt{(2-(-1))^{2}+(-1-(-5))^{2}+(2-(-10))^{2}}

=\sqrt{(2+1)^{2}+(-1+5)^{2}+(2+10)^{2}}

=\sqrt{(3)^{2}+(4)^{2}+(12)^{2}}

=\sqrt{9+16+144}

=\sqrt{169}=13 unit

Posted by

infoexpert21

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question