Ncert Solutions Maths class 12th
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New answer posted
a month agoContributor-Level 9
A (-3, -6,1), B (2,4, -3). Point P divides AB in ratio k:1.
P = [ (2k-3)/ (k+1), (4k-6)/ (k+1), (-3k+1)/ (k+1)]
P lies on the plane lx + my + nz = 0.
l (2k-3) + m (4k-6) + n (-3k+1) = 0
k (2l + 4m - 3n) = 3l + 6m - n
⇒ k = (3l + 6m - n) / (2l + 4m - 3n)
Plane contains the line (x-1)/-1 = (y+4)/2 = (z+2)/3.
The plane passes through (1, -4, -2) and its normal is perpendicular to the line's direction vector.
-l + 2m + 3n = 0
l (1) + m (-4) + n (-2) = 0 ⇒ l - 4m - 2n = 0
Solving these gives l/-8 = m/-1 = n/-2. Let l=8, m=1, n=2.
k = (3 (8) + 6 (1) - 2) / (2 (8) + 4 (1) - 3 (2) = (24+6-2)/ (16+4-6) = 28/14 = 2.
New answer posted
a month agoContributor-Level 9
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 point on it (2r, 4 - 2r, 5 + 4r) lies on plane lx + my + nz = 2x - 2y + 4z = 0.
⇒ 4r - 2 (4 - 2r) + 4 (5 + 4r) = 0 ⇒ 24r + 12 = 0 ⇒ r = -1/2.
∴ D (-1, 5, 3), C (0, -2, -1).
∴ CD = √ (0 - (-1)² + (-2 - 5)² + (-1 - 3)²) = √ (1 + 49 + 16) = √66
New answer posted
a month agoContributor-Level 10
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) = (4, 3, 2).
New answer posted
a month agoContributor-Level 9
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 coplanar with lines PR and QS.
⇒ TA ⊥ OT
|OT| = √166
|TA| = √5
|OA| = √ (|OT|² + |TA|²) = √171
New answer posted
a month agoContributor-Level 10
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 (ii): a² = 5 - 1 = 4. Since a < 0, a = -2.
The equation of the circle is x² + y² - 2x - 4y - 1 = 0.
Completing the square: (x-1)² + (y-2)² = 1+4+1 = 6.
The center is (1,2) and radius is √6.
The tangent is perpendicular to the line x + 2y = 0 (slope -1/2).
So, the slope of the tangent is 2.
Equation of the tangent: (y-2) = 2 (x-1)
New answer posted
a month agoContributor-Level 10
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 k |
| 2 3 4 |
| 3 4 5 |
= I (15-16) - j (10-12) + k (8-9) = -i + 2j - k
D = | (i + 2j + k) ⋅ (-i + 2j - k)| / √ (-1)² + 2² + (-1)²)
= |-1 + 4 - 1| / √ (1 + 4 + 1)
= 2 / √6
New answer posted
a month agoContributor-Level 10
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
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