Ncert Solutions Maths class 11th

Let x = m(a + λb).
Given m(a + λb) ⋅ (3i + 2j - k) = 0, which leads to λ = -3/8.

The projection of vector x on vector a is given by x ⋅ â, where â is the unit vector of a.
Projection = (x ⋅ a) / |a| = 17√6 / 2
x ⋅ a = (m(a + λb)) ⋅ a = m(a ⋅ a + λ(b ⋅ a)) = m(|a|^2 + λ(b ⋅ a))

The provided text simplifies this to:
m(6 - 3/8 * (-1)) = 17√6 / 2
m * (51/8) = 17 * 6 / 2 (The text seems to have a typo 17x6/2 instead of 17√6 / 2)
Assuming it is 17 * 6 / 2, m * 51/8 = 51, so m = 8.

x = 8(a + (-3/8)b) = 8a - 3b
x = 8( (13/8)i - (14/8)j + (11/8)k ) (The vectors a and b are not fully defined in the provided text)
The final vector is given as x = 13i - 14j + 11k.

|x|^2 = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486.

If the orthocenter and circumcenter of a triangle both lie on the y-axis, the centroid also lies on the y-axis.
The x-coordinate of the centroid is (x1 + x2 + x3) / 3. If the vertices are (cos α, sin α), (cos β, sin β), (cos γ, sin γ), then the x-coordinate of the centroid is (cos α + cos β + cos γ) / 3.
Since the centroid lies on the y-axis, its x-coordinate is 0.
cos α + cos β + cos γ = 0

Using the identity: If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.
Let a = cos α, b = cos β, c = cos γ.
Then, cos^3 α + cos^3 β + cos^3 γ = 3 * cos α * cos β * cos γ.

We need to evaluate the expression:
(cos 3α + cos 3β + cos 3γ) / (cos α * cos β * cos γ)

Using the identity cos 3θ = 4cos^3 θ - 3cos θ:
cos 3α + cos 3β + cos 3γ = (4cos^3 α - 3cos α) + (4cos^3 β - 3cos β) + (4cos^3 γ - 3cos γ)
= 4(cos^3 α + cos^3 β + cos^3 γ) - 3(cos α + cos β + cos γ)

Substitute the known values:
= 4(3 * cos α * cos β * cos γ) - 3(0)
= 12 * cos α * cos β * cos γ

So, the expression becomes:
(12 * cos α * cos β * cos γ) / (cos α * cos β * cos γ) = 12.
The value of the expression squared is 12^2 = 144.

I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the integer period of sin(2πx).
In the interval (0, 1/2), sin(2πx) is between 0 and 1, so [sin(2πx)] = 0.
In the interval (1/2, 1), sin(2πx) is between -1 and 0, so [sin(2πx)] = -1.

At x=0, 1/2, 1, the value is 0.
So, ∫[0 to 1] f(x) dx = ∫[0 to 1/2] 0 dx + ∫[1/2 to 1] -1/e^x dx
= 0 + [-e^(-x) * (-1)] from 1/2 to 1 = [e^(-x)] from 1/2 to 1 = e?¹ - e?¹/².
The total integral is 10 * (e?¹ - e?¹/²).
The text calculates: 10 * ∫[1/2 to 1] (-1)e^(-x) dx = -10[-e^(-x)] from 1/2 to 1 = 10(e?¹ - e?¹/²).
This seems to lead to α = 10, β = -10, γ = 0. Then α+β+γ = 0. The calculation in the text has a sign error in the final result.

Parabola: y² = 4x - 20 = 4(x - 5). Vertex at (5,0).
Line: The text seems to derive the tangent equation y = x - 4. This is not a tangent to the given parabola. The standard tangent to y²=4aX is Y=mX+a/m. Here X=x-5, a=1. So y = m(x-5)+1/m.
The other curve is an ellipse: x²/a² + y²/b² = 1.
The text says x²/2 + (x-4)²/b² = 1. This assumes a² = 2.
x²/2 + (x²-8x+16)/b² = 1
x²(1/2 + 1/b²) - (8/b²)x + (16/b² - 1) = 0.
For tangency, the discriminant (D) of this quadratic equation must be zero.
D = (8/b²)² - 4(1/2 + 1/b²)(16/b² - 1) = 0.
64/b? - 4(8/b² - 1/2 + 16/b? - 1/b²) = 0.
16/b? - (7/b² - 1/2 + 16/b?) = 0.
-7/b² + 1/2 = 0 ⇒ b² = 14.
The calculation in the text is different: 64/b? - 4(1/2 + 1/b²)(16/b² - 1) = 0 implies b=14. This should be b²=14.

• Given vectors OP = xi + yj - k and OQ = -i + 2j + 3xk.
• PQ = OQ - OP = (-1 - x)i + (2 - y)j + (3x + 1)k
• Given |PQ| = √20, so |PQ|² = 20.
  (-1 - x)² + (2 - y)² + (3x + 1)² = 20
  (1 + x)² + (2 - y)² + (3x + 1)² = 20 .(i)
• • Given OP ⊥ OQ, so OP · OQ = 0.
    (x)(-1) + (y)(2) + (-1)(3x) = 0
    -x + 2y - 3x = 0 ⇒ -4x + 2y = 0 ⇒ y = 2x .(ii)

  Substitute (ii) into (i):
  (1 + x)² + (2 - 2x)² + (3x + 1)² = 20
  1 + 2x + x² + 4 - 8x + 4x² + 9x² + 6x + 1 = 20
  14x² = 14 ⇒ x² = 1 ⇒ x = ±1.
  When x = 1, y = 2. When x = -1, y = -2.
  So, (x, y) can be (1, 2) or (-1, -2).

  • Given that OR, OP, and OQ are coplanar, their scalar triple product is 0: [OR OP OQ] = 0.
    Let OR = 3i + j + (z-7)k.
    The determinant is:
    | 3 z-7 -7 |
    | x y -1 | = 0
    | -1 2 3x |

  3(3xy + 2) - (z-7)(3x² + 1) + (-7)(2x + y) = 0 .(iii)

  Putting x = 1 and y = 2 into (iii):
  3(3(1)(2) + 2) + z(1 - 3(1)²) - 7(2(1) + 2) = 0
  3(8) + z(-2) - 7(4) = 0
  24 - 2z - 28 = 0 ⇒ -2z = 4 ⇒ z = -2.

  Then, x² + y² + z² = 1² + 2² + (-2)² = 1 + 4 + 4 = 9.