Ncert Solutions Maths class 12th
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If for a > 0, the feet of perpendiculars from the points A(a, -2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, -a, -1) and D respectively, then the length of line segment CD is equal to :
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If for a > 0, the feet of perpendiculars from the points A(a, -2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, -a, -1) and D respectively, then the length of line segment CD is equal to :
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AC is collinear with l, m, n.
(l/a) = (m/-a) = (n/4) = λ.
Again (0, -a, -1) lies on lx + my + nz = 0.
l (0) + m (-a) + n (-1) = 0 ⇒ -am - n = 0
-a (-aλ) - (4λ) = 0 ⇒ a²λ - 4λ = 0 ⇒ a² = 4, a = ±2.
for a > 0, a = 2. Direction ratios of BD are 2, -2, 4.
Equation of BD: x/2 = (y - 4)/-2 = (z - 5)/4 = r.
Any
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Let P be a plane passing through the points (2,1,0), (4,1,1) and (5,0,1) and R be any point (2,1,6). Then the image of R in the plane P is:
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Let P be a plane passing through the points (2,1,0), (4,1,1) and (5,0,1) and R be any point (2,1,6). Then the image of R in the plane P is:
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The equation of the line is (x-2)/1 = (y-1)/1 = (z-6)/-2.
Let this be equal to k. So, a point on the line is (k+2, k+1, -2k+6).
This point lies on the plane x + y - 2z = 3.
(k+2) + (k+1) - 2 (-2k+6) = 3
2k + 3 + 4k - 12 = 3
6k - 9 = 3
6k = 12 ⇒ k = 2.
The point of intersection is (2+2, 2+1, -2 (2)+6) = (
Upvote Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k respectively. Let R and S be two points such that the direction rations of lines PR and QS are (4, -1, 2) and (-2, 1, -2), respectively. Let lines PR and QS intersect at T. If the vector TA is perpendicular to both PR and QS and the length of vector TA is √5 units, then the modulus of a position vector of A is :
Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k respectively. Let R and S be two points such that the direction rations of lines PR and QS are (4, -1, 2) and (-2, 1, -2), respectively. Let lines PR and QS intersect at T. If the vector TA is perpendicular to both PR and QS and the length of vector TA is √5 units, then the modulus of a position vector of A is :
PR (line): r = (3i - j + 2k) + λ (4i - j + 2k) - (I)
QS (line): r = (i + 2j - 4k) + μ (-2i + j - 2k) - (II)
If they intersect at T then:
3 + 4λ = 1 - 2μ
-1 - λ = 2 + μ
2 + 2λ = -4 - 2μ
Solving the first two equations gives λ = 2 & μ = -5. These values satisfy the third equation.
∴ T (11, -3, 6)
Also, OT is
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Let A = {2,3,4,5,.,30} and '~' be an equivalence relation on A×A, defined by (a,b) ~ (c,d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
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Let A = {2,3,4,5,.,30} and '~' be an equivalence relation on A×A, defined by (a,b) ~ (c,d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
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The equation of the circle is x²+y²+ax+2ay+c=0, with a<0.
x-intercept = 2√ (g² - c) = 2√ (a/2)² - c) = 2√2. So, a²/4 - c = 2 => a² = 8 + 4c - (i)
y-intercept = 2√ (f² - c) = 2√ (a² - c) = 2√5. So, a² - c = 5 => a² = 5 + c - (ii)
Equating (i) and (ii): 8 + 4c = 5 + c => 3c = -3 => c = -1.
Substituting c in
Upvote If the foot of the perpendicular from point (4, 3, 8) on the line L₁ : (x-a)/l = (y-2)/3 = (z-b)/4, l ≠ 0 is (3, 5, 7) then the shortest distance between the line L₁ and line L₂ : (x-2)/3 = (y-4)/4 = (z-5)/5 is equal to:
If the foot of the perpendicular from point (4, 3, 8) on the line L₁ : (x-a)/l = (y-2)/3 = (z-b)/4, l ≠ 0 is (3, 5, 7) then the shortest distance between the line L₁ and line L₂ : (x-2)/3 = (y-4)/4 = (z-5)/5 is equal to:
The shortest distance D between two skew lines is given by the formula:
D = | (a? - a? ) ⋅ (b? x b? )| / |b? x b? |
Line L? : (x-1)/2 = (y-2)/3 = (z-4)/4
Line L? : (x-2)/3 = (y-4)/4 = (z-5)/5
Here, a? = I + 2j + 4k, b? = 2i + 3j + 4k
a? = 2i + 4j + 5k, b? = 3i + 4j + 5k
a? - a? = I + 2j + k
b? x b? = | I j
Upvote If (x,y,z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
3 + (x-11)/((y-19)²(z-12)²) + (y-9)/((x-11)²(z-12)²) + (z-12)/((x-11)²(y-19)²) - (x+y+z)/(14(x-11)(y-19)(z-12))
is equal to :
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If (x,y,z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
3 + (x-11)/((y-19)²(z-12)²) + (y-9)/((x-11)²(z-12)²) + (z-12)/((x-11)²(y-19)²) - (x+y+z)/(14(x-11)(y-19)(z-12))
is equal to :
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The equation of the plane is given as x + y + z = 42. It is also mentioned that x³ + y³ + z³ = 3xyz.
From the identity, if x³ + y³ + z³ - 3xyz = 0, then x + y + z = 0 or x = y = z.
Given the expression:
3 + (x³ + y³ + z³ - 3xyz) / (xyz)²
Since x³ + y³ + z³ = 3xyz, the expression simplifies to:
3 + 0 = 3
Upvote If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x3) + xg(x3) is divisible by x2 + x + 1, then P(1) is equal to.
If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x3) + xg(x3) is divisible by x2 + x + 1, then P(1) is equal to.
P (x) = f (x³) + xg (x³) is divisible by x²+x+1. The roots of x²+x+1=0 are the complex cube roots of unity, ω and ω².
P (ω) = f (ω³) + ωg (ω³) = f (1) + ωg (1) = 0 — (I)
P (ω²) = f (ω²)³) + ω²g (ω²)³) = f (1) + ω²g (1) = 0 — (II)
Subtracting (II) from (I): (ω - ω²)g (1) = 0. Since ω ≠ ω², we must have
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