A space vector makes the angles 150° and 60° with the positive direction of x-and y-axes. The angle made by the vector with the positive direction z-axis is
90°
60°
180°
120°
If is the acute angle between the curves : x2 + y2 = 4x and x2 + y2 = 8 at (2, 2), then a is equal to
1
0
Equation of the plane perpendicular to the line and passing through the point (2, 3, 4) is
2x + 3y + z = 17
x + 2y + 3z = 9
3x + 2y + z = 16
x + 2y + 3z = 20
D.
x + 2y + 3z = 20
Equation of plane passing through (2, 3, 4) is
a(x - 2) + b(y - 3) + c(z - 4) = 0 ...(i)
Since, above plane is perpendicular to the line
Thus, normal to the plane is parallel to the line
So, DR's of normal is (1, 2, 3),
i.e., (a, b, c) = (1, 2, 3)
Now, from Eq (i),
1(x - 2) + 2(y - 3) + 3(z - 4) = 0
The line is parallel to the plane
2x + 3y + 4z = 0
3x + 4y + 5z = 7
2x + y - 2z = 0
x + y + z = 2
If a = and the angle between a and b is , then the area of the triangle formed by these two vectors as two sides is
15