Animated Solution for Mathematics - Three Dimensional Geometry: Let A be vector parallel to line of intersection of planes P1 and P2. Plane P1 is parallel to the vectors 2j^+3k^ and 4j^−3k^ and that P2 is parallel to j^−k^ and 3i^+3j^, then the angle between vector A and a given vector 2i^+j^−2k^ is
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Visualized Solution
Visualizing the Geometry
Plane P1 is parallel to 2j^+3k^ and 4j^−3k^.
Plane P2 is parallel to j^−k^ and 3i^+3j^.
Goal: Find the angle between intersection vector A and given vector B.
Normal to Plane P1
A plane's normal n1 is perpendicular to any two non-collinear vectors on it.
n1=(2j^+3k^)×(4j^−3k^)
Calculating n1
n1=i^00j^24k^3−3
n1=i^(−6−12)−j^(0)+k^(0)
n1=−18i^
Normal to Plane P2
Similarly, for plane P2, the normal n2 is the cross product of its parallel vectors.
n2=(j^−k^)×(3i^+3j^)
Calculating n2
n2=i^03j^13k^−10
n2=i^(0−(−3))−j^(0−(−3))+k^(0−3)
n2=3i^−3j^−3k^
Direction of Intersection A
The line of intersection lies in both planes.
Therefore, vector A is perpendicular to both n1 and n2.
A∥(n1×n2)
Calculating Vector A
A=(−18i^)×(3i^−3j^−3k^)
A=−18(3)(i^×i^)−18(−3)(i^×j^)−18(−3)(i^×k^)
A=0+54k^+54(−j^)=−54(j^−k^)
Angle with Vector B
Let the direction of A be j^−k^.
Given vector B=2i^+j^−2k^.
Formula: cosθ=∣A∣∣B∣A⋅B
Calculating Dot Product and Magnitudes
A⋅B=(0)(2)+(1)(1)+(−1)(−2)=3
∣A∣=02+12+(−1)2=2
∣B∣=22+12+(−2)2=9=3
Solving for cosθ
cosθ=±323
cosθ=±21
Note: We use ± because the line of intersection can be directed either way.
Final Answer
If cosθ=21, then θ=4π.
If cosθ=−21, then θ=43π.
Both 4π and 43π are correct possible angles.
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The Sigma Insight: Equation of a Plane
Solution Diagram
The Geometry of Intersecting Planes
Imagine you are standing in a vast, empty room. In front of you, two massive, flat sheets of glass intersect, forming a single, sharp line where they meet. This is the geometric reality of our problem.
We are given two planes, P1 and P2, and we need to find the angle between their line of intersection—represented by vector A—and a third, external vector B=2i^+j^−2k^. This isn't just algebra; it's spatial navigation.
Phase 1
Unlocking the Normals
To understand a plane, we must understand its normal vector. Think of the normal vector as a compass needle pointing straight out from the surface.
For plane P1, we are given two vectors parallel to it: 2j^+3k^ and 4j^−3k^. To find the normal n1, we take their cross product:
n1=(2j^+3k^)×(4j^−3k^)=−18i^
Now, we repeat this for plane P2, which is parallel to j^−k^ and 3i^+3j^. Calculating n2=(j^−k^)×(3i^+3j^), we arrive at:
n2=3i^−3j^−3k^
Phase 2
The Intersection Line
The line of intersection lies in both planes. This means the direction vector A of this line must be perpendicular to both n1 and n2.
To find a vector perpendicular to two others, we use the cross product:
A=n1×n2=(−18i^)×(3i^−3j^−3k^)
As we expand this, the i^×i^ term vanishes, leaving us with A=−54(j^−k^). Since we only care about the direction, we can simplify this to:
A=j^−k^
Phase 3
The Final Angle
We have our intersection vector A=j^−k^ and our target vector B=2i^+j^−2k^. To find the angle θ between them, we use the dot product formula:
cosθ=∣A∣∣B∣A⋅B
The dot product A⋅B is (0)(2)+(1)(1)+(−1)(−2)=3. The magnitude ∣A∣ is 02+12+(−1)2=2, and the magnitude ∣B∣ is 22+12+(−2)2=3.
Plugging these values into our formula:
cosθ=323=21
Because the line of intersection can be oriented in two directions, we must also consider cosθ=−21. This gives us two possible angles:
θ=4πorθ=43π
Both are correct, reflecting the two ways you can walk along that line of intersection. You have mastered the geometry!
