Quadrilateral pyramid

Question

In the quadrilateral pyramid P-ABCD, the base surface ABCD is rectangular, the base surface ABCD 丄 PA, AB= √3, BC=1, PA=2, and E is the midpoint of PD.

1) calculate the cosine of the Angle between AC and PB

2)find a point N in the side PAB, make the NEL plane PAC, and find the coordinates of point N.

Answer 1

Given:

1. Pyramid P−ABCDP-ABCDP−ABCD with rectangular base ABCDABCDABCD.
2. AB=3AB = \sqrt{3}AB=3​, BC=1BC = 1BC=1, PA=2PA = 2PA=2.
3. EEE is the midpoint of PDPDPD.

To find:

1. The cosine of the angle between ACACAC and PBPBPB.

Solution:

Since ABCDABCDABCD is a rectangular base and PAPAPA is given as 2 units, we can position the pyramid in a coordinate system.

Let's assume:

1. A=(0,0,0)A = (0, 0, 0)A=(0,0,0)
2. B=(3,0,0)B = (\sqrt{3}, 0, 0)B=(3​,0,0)
3. C=(3,1,0)C = (\sqrt{3}, 1, 0)C=(3​,1,0)
4. D=(0,1,0)D = (0, 1, 0)D=(0,1,0)
5. P=(0,0,h)P = (0, 0, h)P=(0,0,h) (where h=PA=2h = PA = 2h=PA=2)

Now, EEE is the midpoint of PDPDPD:

1. D=(0,1,0)D = (0, 1, 0)D=(0,1,0)
2. P=(0,0,2)P = (0, 0, 2)P=(0,0,2)
3. EEE (midpoint of PDPDPD) =(0+02,1+02,0+22)=(0,0.5,1)= \left( \frac{0+0}{2}, \frac{1+0}{2}, \frac{0+2}{2} \right) = (0, 0.5, 1)=(20+0​,21+0​,20+2​)=(0,0.5,1)

Now, find vectors ACACAC and PBPBPB:

1. AC=C−A=(3,1,0)AC = C - A = (\sqrt{3}, 1, 0)AC=C−A=(3​,1,0)
2. PB=B−P=(3,0,−h)=(3,0,−2)PB = B - P = (\sqrt{3}, 0, -h) = (\sqrt{3}, 0, -2)PB=B−P=(3​,0,−h)=(3​,0,−2)

To find the cosine of the angle θ\thetaθ between ACACAC and PBPBPB: cos⁡θ=AC⋅PB∥AC∥∥PB∥\cos \theta = \frac{AC \cdot PB}{\|AC\| \|PB\|}cosθ=∥AC∥∥PB∥AC⋅PB​

Calculate the dot product AC⋅PBAC \cdot PBAC⋅PB: AC⋅PB=3⋅3+1⋅0+0⋅(−2)=3AC \cdot PB = \sqrt{3} \cdot \sqrt{3} + 1 \cdot 0 + 0 \cdot (-2) = 3AC⋅PB=3​⋅3​+1⋅0+0⋅(−2)=3

Calculate the magnitudes: ∥AC∥=(3)2+12=4=2\|AC\| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2∥AC∥=(3​)2+12​=4​=2 ∥PB∥=(3)2+02+(−2)2=7\|PB\| = \sqrt{(\sqrt{3})^2 + 0^2 + (-2)^2} = \sqrt{7}∥PB∥=(3​)2+02+(−2)2​=7​

Now, compute cos⁡θ\cos \thetacosθ: cos⁡θ=327=327\cos \theta = \frac{3}{2 \sqrt{7}} = \frac{3}{2\sqrt{7}}cosθ=27​3​=27​3​

Therefore, the cosine of the angle between ACACAC and PBPBPB is 327\frac{3}{2\sqrt{7}}27​3​.

To find a point NNN in the triangle PABPABPAB such that the plane △PAC\triangle PAC△PAC passes through NNN, and then find the coordinates of NNN:

Solution:

Firstly, determine the coordinates of point LLL:

1. LLL is the midpoint of PCPCPC.
2. P=(0,0,2)P = (0, 0, 2)P=(0,0,2)
3. C=(3,1,0)C = (\sqrt{3}, 1, 0)C=(3​,1,0)
4. L=(0+32,0+12,2+02)=(32,12,1)L = \left( \frac{0+\sqrt{3}}{2}, \frac{0+1}{2}, \frac{2+0}{2} \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2}, 1 \right)L=(20+3​​,20+1​,22+0​)=(23​​,21​,1)

Now, find NNN such that the plane △PAC\triangle PAC△PAC passes through NNN:

1. NNN lies in △PAB\triangle PAB△PAB, which is on the base plane z=0z = 0z=0.
2. A convenient point on △PAB\triangle PAB△PAB could be A=(0,0,0)A = (0, 0, 0)A=(0,0,0).

Therefore, N=A=(0,0,0)N = A = (0, 0, 0)N=A=(0,0,0).

So, NNN is (0,0,0)(0, 0, 0)(0,0,0).

Summary of Coordinates:

1. N=(0,0,0)N = (0, 0, 0)N=(0,0,0)

This point NNN lies in the plane △PAC\triangle PAC△PAC, which is the plane passing through points P=(0,0,2)P = (0, 0, 2)P=(0,0,2), A=(0,0,0)A = (0, 0, 0)A=(0,0,0), and C=(3,1,0)C = (\sqrt{3}, 1, 0)C=(3​,1,0).

Thus, the coordinates of point NNN are (0,0,0)\boxed{(0, 0, 0)}(0,0,0)​.