Let a be a vector parallel to the line of intersection of the planes P1 and P2- Plane P1 is parallel to the vectors 2- -3 k- and 4- -3 k- Plane P2 is parallel to the vectors - - k- and 3- -3 - Then the angle between the vector a and a given vector 2- - 2k- can be

Normal vector to P_1 is, n_1=(2ĵ +3 k̂)×(4ĵ 3 k̂)= 18 î Normal vector to P_2 is, n_2=(ĵ k̂)×(3î +3 ĵ)= 3 ( î ĵ k̂) Now, a is parallel to nbsp;n_1× nbsp; ...

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Byju's Answer

Standard IX

Mathematics

Equation of a Plane Passing through a Point and Perpendicular to a Given Vector

Let a be a ve...

Question

Let $\to a$ be a vector parallel to the line of intersection of the planes $P_{1}$ and $P_{2} .$ Plane $P_{1}$ is parallel to the vectors $2^j + 3^k$ and $4^j - 3^k .$ Plane $P_{2}$ is parallel to the vectors $^j -^k$ and $3^i + 3^j .$ Then the angle between the vector $\to a$ and a given vector $2^i +^j - 2^k$ can be

\frac{π}{4}

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\frac{π}{3}

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\frac{2 π}{3}

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\frac{3 π}{4}

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Solution

The correct option is D $\frac{3 π}{4}$
Normal vector to $P_{1}$ is,
$_{1} = (2^j + 3^k) \times (4^j - 3^k) = - 18^i$
Normal vector to $P_{2}$ is,
$_{2} = (^j -^k) \times (3^i + 3^j) = 3 (^i -^j -^k)$

Now, $\to a$ is parallel to $_{1} \times_{2} .$
So, angle between $\to a$ and $2^i +^j - 2^k$ is same as angle between $_{1} \times_{2}$ and $2^i +^j - 2^k .$
$_{1} \times_{2} = 54 (-^j +^k)$

Let $θ$ be the angle between $_{1} \times_{2}$ and $2^i +^j - 2^k .$
Then, $cos θ = \frac{54 (- 1 - 2)}{54 \cdot \sqrt{2} \cdot \sqrt{9}} = - \frac{1}{\sqrt{2}}$
$\Rightarrow θ = \frac{3 π}{4}$
Angle depends upon the direction of $\to a$ which is parallel to $_{1} \times_{2} .$
If $\to a$ is anti-parallel to $_{1} \times_{2},$ then $θ = \frac{π}{4}$

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Let →a be a vector parallel to the line of intersection of the planes P1 and P2. Plane P1 is parallel to the vectors 2^j+3^k and 4^j−3^k. Plane P2 is parallel to the vectors ^j−^k and 3^i+3^j. Then the angle between the vector →a and a given vector 2^i+^j−2^k can be