Angle between the planes x-y-2 z-6 and 2 x-y-z-9 is-A- -4B- -6C- -3D- -2

Angle between the planes= angle between the normals The normal vectors are, n_1 = î + ĵ+ 2k̂ n_2 = 2î ĵ +k̂ Taking dot product, n_1 ·n_2 = 2 1 + 2 = 3 ⇒ |n_ ...

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Standard XII

Physics

Rotational EMF

Angle between...

Question

Angle between the planes $x + y + 2 z = 6$ and $2 x - y + z = 9$ is,

\frac{π}{4}

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\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is C $\frac{π}{3}$
Angle between the planes= angle between the normals
The normal vectors are,
${\to n}_{1} =^i +^j + 2^k {\to n}_{2} = 2^i -^j +^k$
Taking dot product,
${\to n}_{1} \cdot {\to n}_{2} = 2 - 1 + 2 = 3 \Rightarrow | {\to n}_{1} | | {\to n}_{2} | = 3 \Rightarrow \sqrt{6} \times \sqrt{6} cos θ = 3 \Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

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Angle between the planes x+y+2z=6 and 2x−y+z=9 is,

The correct option is C π3Angle between the planes= angle between the normals The normal vectors are, →n1=^i+^j+2^k→n2=2^i−^j+^k Taking dot product, →n1⋅→n2=2−1+2=3⇒|→n1||→n2|=3⇒√6×√6cosθ=3⇒cosθ=12⇒θ=π3

Angle between the planes $x + y + 2 z = 6$ and $2 x - y + z = 9$ is,