Class 12 Three Dimensional Geometry · Case Study4 Marksline_applications
Class 12 Maths · Three Dimensional Geometry · Case Study #94
A moving object follows x=3t,y=−4t,z=t, in kilometres, for t≥0 seconds. The xy-plane is ground level. For part (iii)(a), consider independently a point P=(5,−8,10). Answer (i),(ii), and either (iii)(a) or (iii)(b).
Part (1) 1 Marks
Describe the path.
Answer: The ray r=t(3,−4,1), t≥0. Its supporting line is x/3=y/(−4)=z.
- The position is t times the fixed direction (3,−4,1).
- Thus the physical path is the ray t≥0, on the stated straight line.
Part (2) 1 Marks
Find displacement magnitude after 5 seconds.
Answer: 5√26 km.
- At t=5, the point is (15,−20,5).
- Its displacement magnitude is √(225+400+25)=5√26 km.
Part (3) 2 Marks
Find the height of the independent point P=(5,−8,10) above the xy-plane.
Answer: 10 km.
- The height above z=0 is |z|.
- The perpendicular foot of P is (5,−8,0); its vertical separation is |10−0|.
- The height is 10 km.
Part (4) 2 Marks
Find k so that directions (2,3,k) and (−2,−1,7) are orthogonal.
Answer: k=1.
- Set the dot product of the two directions to zero.
- −4−3+7k=0, giving 7k=7.
- Thus k=1.