3 The figure below shows a right pyramid VABCD with a square

3. The figure below shows a right pyramid VABCD with a square base ABCD, standing horizontally on a cuboid ABCDEFGH. It is given that VA VB VC VD 5 cm. EF FG = 4 cm and AE-2 cm, as shown in the diagram, O is the cent re of the square base EFGH. The unit vectors i, j, k are parallel to EF, FG, EA respectively. 5 2 4 (a) (3 marks) Find the height, OV, of the figure. Give your answer in exact fornm (b) (3 marks) Find (c) (3 marks) A plane has the equation 3x-2y+az = . The planes , VBC and VAB Interpret your answer geometrically have no point in common. Find and . Give your answers in exact form

Solution

1 let us draw OV such that it cuts the bar of prism at S.

Thus OV=OS+SV

Clearly, OS=AE=2

SV is the altitude of triangle BVD.

In triangle BVD,

BD will be equal to the diagonal of the square Base ABCD.

This will be equal to 4*(2)^0.5

Thus,BD=5.66

S is mid point of BD

Thus, SD=5.66/2=2.83

SV is the altitude of triangle SVD which can be found by using pythagorean therum.

Thus, SV^2 =DV^2 -SD^2

Keeping values

SV^2 =5^2 -2.83^2

Thus,

SV=4.12

Thus,OV=2+4.12=6.12