If the angle between the planes r → .( m i ^ − j ^ +2 k ^ )+3=0 and r → .( 2 i ^ − m j ^ − i ^ )−5=0 is (π)/3 then m=

Correct answer: (C) 3. Angle between given planes is Cos θ=(n-_1 ⋅ n-_2)/(n-_1 n-_2) ⇒1/2=(2m+m-2)/(√(m^2 + 5) √(m^2 + 5)) 1/2=(3m-2)/(m^2+5)⇒m^2+5=± 6m-4 ⇒ m^2+5=6m-4,m^2+5=-6m+4 m^2-6m+9=0,m^2+6m+1=0 m-3^2=0 m=3

Q70 / 150 · MHT CET 2017 · Mathematics

If the angle between the planes $\vec{r} . (m \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} . (2 \hat{i} - m \hat{j} - \hat{i}) - 5 = 0$ is $\frac{π}{3}$ then $m =$

A
$2$
B
$\pm 3$
Correct answer
C
$3$
D
$- 2$

Solution

Angle between given planes is
Cos

θ = |\frac{{\bar{n}}_{1} \cdot {\bar{n}}_{2}}{|{\bar{n}}_{1}| |{\bar{n}}_{2}|}| \Rightarrow \frac{1}{2} = |\frac{2 m + m - 2}{\sqrt{m^{2} + 5} \sqrt{m^{2} + 5}}|

\frac{1}{2} = |\frac{3 m - 2}{m^{2} + 5}| \Rightarrow m^{2} + 5 = \pm (6 m - 4)

\Rightarrow m^{2} + 5 = 6 m - 4, m^{2} + 5 = - 6 m + 4

m^{2} - 6 m + 9 = 0, m^{2} + 6 m + 1 = 0

{(m - 3)}^{2} = 0

m = 3

Hence, the correct answer is option: 'C'.

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If the angle between the planes r → .( m i ^ − j ^ +2 k ^ )+3=0 and r → .( 2 i ^ − m j ^ − i ^ )−5=0 is π3 then m=

Solution

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If the angle between the planes $\vec{r} . (m \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} . (2 \hat{i} - m \hat{j} - \hat{i}) - 5 = 0$ is $\frac{π}{3}$ then $m =$