Let A be vector parallel to line of intersection of planes P 1 and P 2 through origin- P 1 is parallel to the vectors 2 -3 k- and 4 -3 k- and P 2 is parallelto -k- and 3 -3 - then the angle between vector A and 2 i - j -2 k- isA- w-2B- n-4C- -4D- 3 -4

The correct options are B π/4 D 3π/4Let vector A⃗O⃗ be parallel to line of planes P_1 and P_2 through origin. Normal to plane p_1 is n⃗_⃗1⃗ = [(2j⃗ + 3k⃗)× (4ĵ ...

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Byju's Answer

Standard IX

Mathematics

Applications of Cross Product

Let A be vect...

Question

Let $\to A$ be vector parallel to line of intersection of planes $P_{1}$ and $P_{2}$ through origin. $P_{1}$ is parallel to the vectors $2^j + 3^k$ and $4^j - 3^k$ and $P_{2}$ is parallel to $^j -^k$ and $3^i + 3^j,$ then the angle between vector $\to A$ and $2 \to i + \to j - 2^k$ is

\frac{π}{2}

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\frac{π}{4}

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\frac{π}{4}

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\frac{3 π}{4}

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Solution

The correct options are
B $\frac{π}{4}$
D $\frac{3 π}{4}$
Let vector $\to A O$ be parallel to line of planes $P_{1}$ and $P_{2}$ through origin.
Normal to plane $p_{1}$ is
$_{1} = [(2 \to j + 3 \to k) \times (4^j - 3^k)] = - 18^i$
Normal to plane $p_{2}$ is
$_{2} = (^j -^k) \times (3^i + 3^j) = 3^i - 3^j - 3^k S o, \to O A i s p a r a l l e l t o \pm (_{1} \times_{2}) = 54^j - 54^k$
$∴ Angle between 54 (^j -^k) a n d (2^i +^j - 2^k) i s c o s θ = \pm (\frac{54 + 108}{3.54. \sqrt{2}}) = \pm \frac{1}{\sqrt{2}} ∴ θ = \frac{π}{4}, \frac{3 π}{4}$
Hence, B. and D. are correct answers.

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Similar questions

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Let →A be vector parallel to line of intersection of planes P1 and P2 through origin. P1 is parallel to the vectors 2^j+3^k and 4^j−3^k and P2 is parallel to ^j−^k and 3^i+3^j, then the angle between vector →A and 2→i+→j−2^k is

Let $\to A$ be vector parallel to line of intersection of planes $P_{1}$ and $P_{2}$ through origin. $P_{1}$ is parallel to the vectors $2^j + 3^k$ and $4^j - 3^k$ and $P_{2}$ is parallel to $^j -^k$ and $3^i + 3^j,$ then the angle between vector $\to A$ and $2 \to i + \to j - 2^k$ is