Let A- be vector parallel to line of intersection of planes P1 and P2 through origin- P1 is parallel to the vector 2-3k- and 4-3k- and P2 is parallel to -k- and 3-3- then the angle between vector A- and 2-2k- is

The correct options are A π2,3π4 C π4,π2 The normal vector of plane #160; P_1 #160;is #160; (2 ĵ + 3k̂) × (4 ĵ 3k̂), #160;i.e. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Equation of a Plane Passing through a Point and Parallel to the Two Given Vectors

Let A⃗ be v...

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 vector $2^j + 3^k$ 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 $2^i +^j - 2^k$ is

\frac{π}{2}, \frac{3 π}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{4}, \frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{6}, \frac{7 π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}, \frac{3 π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $\frac{π}{2}, \frac{3 π}{4}$
C $\frac{π}{4}, \frac{π}{2}$
The normal vector of plane $P_{1}$ is $(2^j + 3^k) \times (4^j - 3^k),$ i.e. $(- 6^i - 12^i) .$

Assuming that point $(0, 2, 3)$ passes through the first plane we have the plane equation as $- 18 x = 0,$ i.e. $x = 0.$

Similarly for $P_{2}$ , we have the normal vector as $(^j -^k) \times (3^i + 3^j) = (- 3^k - 3^j + 3^i)$

The equation of $P_{2}$ thus becomes $3 (x - 0) - 3 (y - 1) - 3 (z + 1) = 0,$ i.e. $x - y - z = 0$

The line of intersection of two planes is passing through $(0, 0, 0)$ and $(0, 1, - 1)$ , i.e. $r = t y - t z .$

So the point $A$ can be $(0, 1, - 1)$ and so $cos θ$ equals $\frac{3}{3 \sqrt{2}} = \frac{1}{\sqrt{2}}$ implies $θ = \frac{π}{4}$

Suggest Corrections

Similar questions

Q. Let

\to A

be vector parallel to line of intersection of planes

P_{1}

and

P_{2}

, Plane

P_{1}

is parallel to the vectors

2^j + 3^k

and

4^j - 3^k

and that

P_{2}

is parallel to

^j -^k

and

3^i + 3^j

then the angle between vector

\to A

and a given vector

2^i +^j - 2^k

Q. 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 vectors

¯ ¯¯ ¯ A

and

2^i +^j - 2^k

Q. 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

Q. Let

\to A

be vector parallel to the line of intersection of planes

p_{1}

and

p_{2}

through the origin.

p_{1}

is parallel to the vectors

\to a = 2^j + 3^k

and

\to b = 4^j - 3^k

and

p_{2}

is parallel to the vectors

\to c =^j -^k

and

\to d = 3^i + 3^j

. The angle between

\to A

and

2^i +^j - 2^k

is:

Q. A non-zero vector

\to a

is parallel to the line of intersection of plane

p_{1}

determined by

^i +^j

and

^i

and plane

P_{2}

determined by vectors

^i -^j

and

^i +^k

, then angle between a and vector

^i - 2^j + 2^k

Join BYJU'S Learning Program

Let →A be vector parallel to line of intersection of planes P1 and P2 through origin. P1 is parallel to the vector 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 vector $2^j + 3^k$ 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 $2^i +^j - 2^k$ is