Three Dimensional Geometry
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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 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
New answer posted
a month agoContributor-Level 10
The equation of a plane is determined by a point it passes through and a normal vector. The plane passes through (1, -6, -5). Its normal vector (a, b, c) is perpendicular to two other vectors, derived from the given equations:
4a - 3b + 7c = 0
3a + 4b + 2c = 0
The direction of the normal vector (a, b, c) is found by the cross product of (4, -3, 7) and (3, 4, 2):
a = (-3) (2) - 7 (4) = -34.
b = 7 (3) - 4 (2) = 13.
c = 4 (4) - (-3) (3) = 25.
So the plane equation is -34 (x-1) + 13 (y+6) + 25 (z+5) = 0.
The point (1, -1, α) lies on this plane:
-34 (1-1) + 13 (-1+6) + 25 (α+5) = 0.
0 + 13 (5) + 25α + 125 = 0.
65 + 25α + 125 = 0 ⇒ 25α = -190
New answer posted
a month agoContributor-Level 10
The vector PQ is given by Q - P = (-3-1, 5-3, 2-a) = (-4, 2, 2-a).
This vector is collinear with 2i - j + k = (2, -1, 1).
This means their components are proportional: -4/2 = 2/ (-1) = (2-a)/1.
From -2 = 2-a, we find a=4.
The midpoint M of PQ is (-3+1)/2, (5+3)/2, (2+4)/2) = (-1, 4, 3).
M lies on the plane 2x - y + z - b = 0.
Substitute the coordinates of M: 2 (-1) - 4 + 3 - b = 0 ⇒ -2 - 4 + 3 = b ⇒ b = -3.
New answer posted
a month agoContributor-Level 10
Given a right-angled triangle with one angle π/3 (or 60°), and the side adjacent to it is h/2.
Using the tangent function: tan (π/3) = opposite / adjacent.
Let the adjacent side be 2. Then tan (π/3) = h/2.
We know tan (π/3) = √3.
So, h/2 = √3, which implies h = 2√3.
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