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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2024 · Mathematics · Three Dimensional Geometry

If l, m, n > are the direction cosines of a normal to the plane 2...

If ⟨l,m,n>\langle l, m, n > are the direction cosines of a normal to the plane 2x−3y+6z+4=02 x-3 y+6 z+4=0, then what is the value of 49(7l2+m2−n2)?49 \left(7 l²+m²-n²\right) ?

Options

  1. A0
  2. B1
  3. C3
  4. D71

Correct answer

B. 1

Step-by-step solution

The plane equation is ax+by+cz=da x+b y+c z=d Given the plane equation is 2x−3y+6z+4=0⇒2x−3y+6z=−42 x-3 y+6 z+4=0\Rightarrow \quad 2 x-3 y+6 z=-4 Comparing then we get, a=2,b=−3,c=6,d=−4∴a2+b2+c2=(2)2+(−3)2+(6)2=4+9+36=49=7\begin{aligned} \mathrm{a}=2, b & =-3, c=6, d=-4 \\ \therefore \quad \sqrt{a^2+b^2+c^2} & =\sqrt{(2)^2+(-3)^2+(6)^2} \\ & =\sqrt{4+9+36}=\sqrt{49}=7 \end{aligned} Direction cosines of normal to the plane l=aa2+b2+c2=27m=ba2+b2+c2=−37n=ca2+b2+c2=67<l,m,n>=⟨27,−37,67⟩∴49(7l2+m2−n2)=49[7(27)2+(−37)2−(67)2]=49[2849+949−3649]=49[149]=1\begin{aligned} l & =\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{7} \\ m & =\frac{b}{\sqrt{a^2+b^2+c^2}}=-\frac{3}{7} \\ n & =\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{6}{7} \\ < l, m, n > & =\left\langle\frac{2}{7},-\frac{3}{7}, \frac{6}{7}\right\rangle \\ \therefore \quad 49\left(7 l^2+m^2-n^2\right) & =49\left[7\left(\frac{2}{7}\right)^2+\left(-\frac{3}{7}\right)^2-\left(\frac{6}{7}\right)^2\right] \\ & =49\left[\frac{28}{49}+\frac{9}{49}-\frac{36}{49}\right] \\ & =49\left[\frac{1}{49}\right]=1 \end{aligned}
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