If the points A(1, 2, 3), B(-1. 1. 2). C(2, 3, 4) and D(-1. x. 0) are coplanar find the value of x.(full solution)
Answers
Answer:
We have A(3,2,1); B(4,x,5); C(4,2,—2) and D(0,5,—1) as the given points. AD now,
AB
= position vector of B — position vector of A
=(4
i
+x
j
+5
k
)−(3
i
+2
j
+
k
)=
i
+(x−2)
j
+4
k
AC
= position vector of C — position vector of A
= (4
i
+2
j
−2
k
)−(3
i
+2
j
+
k
)=
i
+0
j
−3
k
AD
= position vector of D — position vector of A
=(6
i
+5
j
−
k
)−(3
i
+2
j
+
k
)=3
i
+3
j
−2
k
Since, A. B. C and D arc coplanar. then
[
AB
AC
AD
]=0
∣
∣
∣
∣
∣
∣
∣
∣
1
1
3
(x−2)
0
3
4
−3
−2
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒1[0+9]−(x−2)[−2+9]+4[3−0]=0
⇒9−7(x−2)+4×3=0
⇒7x=35
⇒x=5
Step-by-step explanation:
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Therefore, the solution is x = 4
Given: If the points A(1, 2, 3), B(-1. 1. 2). C(2, 3, 4) and D(-1. x. 0) are coplanar
To Find: find the value of x.
Solution:
Two or more points are said to be coplanar if they lie on the same plane. For the four given points to be coplanar, the area of the triangle formed by any three of the points should be zero. Let's take points A, B, and C to form the triangle ABC.
The position vector of point A is:
a = [1, 2, 3]
The position vector of point B is:
b = [-1, 1, 2]
The position vector of point C is:
c = [2, 3, 4]
The cross product of AB and AC will give us the area vector of the triangle ABC:
AB x AC = [-2, -1, -1] x [1, 1, 1] = [-2, 1, 1]
The magnitude of the area vector is:
|AB x AC| = sqrt((-2)^2 + 1^2 + 1^2) = sqrt(6)
Therefore, the area of the triangle ABC is:
Area of triangle ABC = (1/2) |AB x AC| = (1/2) sqrt(6)
Simplifying, we get:
-x + 2y - z + 2 = 0
Now, we substitute the value of point D into this equation and solve for x:
-x + 2y - z + 2 = 0 (equation of the plane containing A, B, and C)
-x + 2x - 0 + 2 = 0 (substituting point D(-1, x, 0))
Therefore, the solution is x = 4
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