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Q1. The signs of the X,Y and Z coordinates of a point that lies in the octant OXYZ' is

• (- , + , +)
• ( + ,  +  , -)
• ( + , - , -)
• (-, - , +)

Solution

Each coordinate of the point P takes the positive or negative sign depending on whether L, M, N lies on the positive or negative X , Y and Z axis.Following this rule , the signs of the  triplet of coordinates  in the 8 octants will be :OXYZ(+, +, +), OX'YZ (- , + , +), OX'Y'Z( - , - , +) OXY'Z (+ , - , +), OXYZ'( +, +, -) , OX'YZ'( -, + , - ), OX'Y'Z'( - , - , -),  OXY'Z'( + , - , -) ,

Q2. Which of the following are not the direction ratios of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2)

• 8,10,-2; 4,4,-6; 4,6,4
• 8,-10,-2; 4,4,-6; 4,6,4
• 4,6,4; -8,10,-2; -4,-4,-6
• 4,-4,-6; 4,6,4; 8,10,-2

Solution

Q3. The length of the perpendicular from the origin to the plane 3x + 2y - 6z = 21 is:

• 3
• 7
• 14
• 21

Solution

The perpendicular distance from the origin to the plane is

Q4. The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y - z+4) =0 is

• x(1 + 2k) + y(1 + 3k) + z(1 - k) + (-1 + 4k) =0
• x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) =0
• x ( 1-2k) + y(1+3k) +z(1-k) +(-1+4k) =0
• x(1+2k) +y(1+3k)+z(1-k) +(-1 - 4k)= 0

Solution

The equation of the plane which passes through the point of intersection of the two given planes is given by x + y + z -1 + k ( 2x + 3y - z + 4) = 0 x(1 + 2k) + y(1 + 3k) + z(1 - k) + (-1 + 4k) =0

Q5. The points (-2, 3, 5), ( 1, 2, 3) and (7, 0, -1) are collinear because.

• AC + BC = AB
• AB + AC = BC
• AB + BC = AC
• none of the above

Solution

Let the given points be A , B and C , such that A(2, 3 , 5 ), B(1, 2, 3) and C(7, 0, -1). Therefore, the points A, B , C are collinear points

Q6. The angle between two lines whose direction ratios are 1,2,1 and 2,-3,4 is:

• 90^o
• 30^o
• 45^o
• 60^o

Solution

Q7. Find direction cosines of y-axis.

• 1,0,0
• 0,1,0
• 0,0,1
• 0,0,0

Solution

The y-axis makes angles 90^o, 0^o and 90^o respectively with x, y and z-axis.Therefore, the direction cosines of x-axis are cos 90^o, cos 0^o, cos 90^o i.e., 0 ,1,0 direction cosines of y-axis are 0, 1, 0

Q8. The equation of the plane passing through the point (3, - 3, 1) and perpendicular to the line joining the points (3, 4, - 1) and (2, - 1, 5) is:

• x + 5y - 6z + 18 = 0
• x - 5y + 6z + 18 = 0
• - x - 5y + 6z + 18 = 0
• - x - 5y - 6z + 18 = 0

Solution

The equation of the plane passing through the point (3,- 3, 1) is: a(x - 3) + b(y + 3) + c(z - 1) = 0 and the direction ratios of the line joining the points (3, 4, - 1) and (2, - 1, 5) is 2 - 3, -1 -4, 5 + 1, i.e., -1, - 5, 6. Since the plane is perpendicular to the line whose direction ratios are 1, - 5, 6, therefore, direction ratios of the normal to the plane is -1, - 5, 6. So, required equation of plane is: -1(x - 3) - 5(y + 3) + 6(z - 1) = 0 i.e. x + 5y - 6z + 18 = 0.  

Q9. The equation of the plane passing through the line of intersection of the planes x-2y+3z+8=0 and 2x-7y+4z-3=0 and the point (3, 1, -2) is:

• 2x-5y+4z+9=0
• 6x-15y+12z+32=0
• 6x-15y+12z+29=0
• 6x-15y+16z+29=0

Solution

Q10. The distance of the point (2, 3, - 5) from the plane x + 2y - 2z = 9 is:

• 2 units
• 3/2 units
• 3 units
• 10/3 units

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