Chapter: Analytical Geometry of Two & Three-Dimensions
Explanation:
Since, B(0, 12/k, 0) lies on sphere
∴ 0 + (12/k)² + 0 + 0 − 3(12/k) − 0 = 0
⇒ (12/k)[12/k − 3] = 0
⇒ k = 4
For the following two (02) items:
A plane P is parallel to the line having direction ratios ⟨1, 3, 2⟩ and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0.
What is the equation of the plane P?
What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratio is (3, 2, 1)?
If p is the perpendicular distance from origin to the plane passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1), then what is 3p2 equal to?
If < l, m, n > are the direction cosines of a normal to the plane 2x – 3y + 6z + 4 = 0, then what is the value of 49(7l2 + m2 – n2)?
Given the circle:
x² + y² + 2x + 6y + 1 = 0
To find the centre (a, b) and radius c, rewrite the equation in the form
(x − h)² + (y − k)² = r²
Completing the square:
(x² + 2x + 1) − 1 + (y² + 6y + 9) − 9 + 1 = 0
⇒ (x + 1)² − 1 + (y + 3)² − 9 + 1 = 0
⇒ (x + 1)² + (y + 3)² − 9 = 0
⇒ (x + 1)² + (y + 3)² = 9
⇒ [x − (−1)]² + [y − (−3)]² = 3²
Therefore, the centre is (−1, −3) and the radius is 3.
∴ a = −1, b = −3, c = 3
∴ a² + b² + c²
= (−1)² + (−3)² + 3²
= 1 + 9 + 9
= 19
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