Three Dimensional Geometry

Direction cosines of a vector making angles with the axes satisfy: Let , then: Try : Now try : Hence, both values satisfy the condition.

Let be the angles made by the ray with the positive X, Y, and Z-axes respectively. The direction cosines of the ray are , , . We know the identity , which means . Given: Angle with Y-axis, . So, . Angle with Z-axis, . So, . Substitute these into the identity: . So, . We need to find the sine of the angle made by the ray with the X-axis, which is . We know . So, . Since : . . The angle made by a ray with an axis is usually taken to be in the range . In this range, . Therefore, . This matches option (a).

Let the plane intersect the X, Y, Z axes at points A, B, C respectively. Let A = , B = , C = . These are the x, y, z intercepts. The equation of the plane in intercept form is . The centroid of has coordinates: . Given that the centroid is . So, . . . The equation of the plane is . To find the perpendicular distance from the origin (0,0,0) to this plane, we can convert the equation to the form . Multiply by the LCM of 6, 9, 15. LCM(6,9,15): , , . LCM = . Multiply by 90: . The perpendicular distance from a point to the plane is given by: Here, the point is origin , and the plane is . So, . We need to find . . So it's between 10 and 20. Numbers ending in 1 when squared end in 1 (e.g., 1, 9). Try . So . This matches option (d).

Equation of plane with intercepts on axes is: Given: Find direction cosines: So direction cosines: