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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Vector Algebra

NDA 2010 · Mathematics · Vector Algebra

P Q R S is a parallelogram, where P Q =3 i +2 j -m k , P S = i +3 j...

PQRSP Q R S is a parallelogram, where PQ→=3i^+2j^−mk^\overrightarrow{P Q}=3 \hat{i}+2 \hat{j}-m \hat{k}, PS→=i^+3j^+k^\overrightarrow{P S}=\hat{i}+3 \hat{j}+\hat{k} and the area of the parallelogram is 90.Whatisthevalueof\sqrt{90}. What is the value ofm??

Options

  1. A1
  2. B-1
  3. C2
  4. D-2

Correct answer

A. 1

Step-by-step solution

 Area of  parallelogram =12∣(PQ→×PS→)∣∴90=∣i^j^k^32−m131∣=[i^(2+3m)−j^(3+m)+k^(9−2)]90=(2+3m)2+(3+m)2+49=4+12m+9m2+9+m2+6m+4990=10m2+18m+62\begin{aligned} & \text{ Area\text{ of } parallelogram }=\frac{1}{2}|(\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PS}})| \\ & \therefore \sqrt{90}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -m \\ 1 & 3 & 1\end{array}\right| \\ & =[\hat{i}(2+3 m)-\hat{j}(3+m)+\hat{k}(9-2)] \\ & \sqrt{90}=\sqrt{(2+3 m)^2+(3+m)^2+49} \\ & =\sqrt{4+12 m+9 m^2+9+m^2+6 m+49} \\ & \sqrt{90}=\sqrt{10 m^2+18 m+62}\end{aligned}90=10m2+18m+62⇒10m2+18m+62−90=0⇒10m2+18m−28=0⇒5m2+9m−14=0⇒5m2−5m+14m−14=05m(m−1)+14(m−1)=0m−145,m=1\begin{aligned} & \sqrt{90}=\sqrt{10 m^2+18 m+62} \\ \Rightarrow & 10 m^2+18 m+62-90=0 \\ \Rightarrow & 10 m^2+18 m-28=0 \\ \Rightarrow & 5 m^2+9 m-14=0 \\ \Rightarrow & 5 m^2-5 m+14 m-14=0 \\ & 5 m(m-1)+14(m-1)=0 \\ & m-\frac{14}{5}, m=1\end{aligned}$
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