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New answer posted
3 months agoContributor-Level 10
a = i + j + 2k
b = -i + 2j + 3k
a + b = 3j + 5k
a . b = -1 + 2 + 6 = 7
a * b = |i, j, k; 1, 2; -1, 2, 3| = -i - 5j + 3k
(a - b) * b) = (a * b) - (b * b) = a * b
(a * (a - b) * b) = a * (a * b) = (a . b)a - (a . a)b = 7a - 6b
. The expression becomes (a + b) * (7a - 6b) * b)
= (a + b) * (7 (a * b)
= 7 [ (a * (a * b) + (b * (a * b) ]
= 7 [ (
New answer posted
3 months agoContributor-Level 10
sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23
New answer posted
3 months agoContributor-Level 10
S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1
New answer posted
3 months agoContributor-Level 10
x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.
New answer posted
3 months agoContributor-Level 10
x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0
New answer posted
3 months agoContributor-Level 10
Equation of tangent of P (2cosθ, sinθ) is
(cosθ)x + (2sinθ)y = 4
Solving equation of tangent with equation of tangents at major axis ends, i.e. x = -2 and x = 2
For point 'B' (at x=-2):
-2cosθ + 2sinθ y = 4 ⇒ y = (2+cosθ)/sinθ
B (-2, (2+cosθ)/sinθ)
For point 'C' (at x=2):
2cosθ + 2sinθ y = 4 ⇒ y = (2-cosθ)/sinθ
C (2, (2-cosθ)/sinθ)
Now BC is the diameter of circle
Equation of circle: (x+2) (x-2) + (y - (2+cosθ)/sinθ) (y - (2-cosθ)/sinθ) = 0
x²-4 + y² - (4/sinθ)y + (4-cos²θ)/sin²θ = 0
Check if (√3, 0) satisfies this:
(√3)²-4 + 0 - 0 + (4-cos²θ)/sin²θ = -1 + (3+sin²θ)/sin²θ = -1 + 3/sin²θ + 1 = 3/sin²
New answer posted
3 months agoContributor-Level 10
ai+aj+ck, i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab
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