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Find the coordinates of a point P where the line through A(3,-4,-5) and B(2,-3,1) crosses the plane, passing through the point L(2,2,1),M(3,0,1) and N(4,-1,0).Also find the ratio in which P divides the line segment AB.
Answers & Comments
Point of intersection is P(1,-2,7).
P externally divides the line segment AB in the ratio 2:1
step-by-step explanation:
The equation of the plane passing through three given points can be given by
Performing elementary row operations R2=>R1-R2 and R3=>R1-R3,
we get
=》
=》
Solving the above determinant, we get
=》(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0
=》(2x-4)+(y-2)+(z-1)=0
=》2x+y+z-7=0
Therefore, the equation of the plane is
2x+y+z-7=0
Now, the equation of the line passing through two given points is
At the point of intersection, these points satisfy the equation of the plane
2x+y+z-7=0.
Putting the values of x, y and z in the equation of the plane, we get the value of a.
Thus, the point of intersection is P(1, -2, 7).
Now, let P divide the line AB in the ratio m:
n.
By the section formula, we have
Hence, P externally divides the line segment AB in the ratio 2:1
Point of intersection is P(1,-2,7).
P externally divides the line segment AB in the ratio 2:1
step-by-step explanation:
The equation of the plane passing through three given points can be given by
\begin{gathered} |x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |x - 3 \: \: \: \: y - 0 \: \: \: z - 1| = 0 \\ | x - 4 \: \: \: \: y - 1 \: \: \: z - 0| \: \: \: \: \: \: \: \: \end{gathered}
∣x−2 y−2z−1∣
∣x−3y−0z−1∣=0
∣x−4y−1z−0∣
Performing elementary row operations R2=>R1-R2 and R3=>R1-R3,
we get
=》
\begin{gathered} |x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |3 - 2 \: \: \: \: 0 - 2 \: \: \: \: \: 0 \: \: \: \: \: \: | = 0 \\ | x - 4 \: \: \: \: y - 1 \: \: \: z - 0| \: \: \: \: \: \: \: \: \end{gathered}
∣x−2 y−2z−1∣
∣3−20−20∣=0
∣x−4y−1z−0∣
=》
\begin{gathered}|x - 2 \: \: \ \: y - 2 \: \: \: z - 1| \: \: \: \: \: \: \: \: \\ |1 \: \: \: \: \: \: - 2 \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: | = 0 \\ | 2 \: \: \: \: \: \: \: - 3 \: \: \: \: \: \: - 1 \: \: \: \: \: \: \: | \: \: \: \: \: \: \: \: \: \end{gathered}
∣x−2 y−2z−1∣
∣1−20∣=0
∣2−3−1∣
Solving the above determinant, we get
=》(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0
=》(2x-4)+(y-2)+(z-1)=0
=》2x+y+z-7=0
Therefore, the equation of the plane is
2x+y+z-7=0
Now, the equation of the line passing through two given points is
\frac{x - 3}{2 - 3} = \frac{y + 4}{ - 3 + 4} = \frac{z + 5}{1 + 5} = \alpha
2−3
x−3
=
−3+4
y+4
=
1+5
z+5
=α
= > \frac{x - 3}{ - 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \alpha=>
−1
x−3
=
1
y+4
=
6
z+5
=α
\begin{gathered} = > x = ( - \alpha + 3) \\ y = ( \alpha - 4) \\ z = (6 \alpha - 5)\end{gathered}
=>x=(−α+3)
y=(α−4)
z=(6α−5)
At the point of intersection, these points satisfy the equation of the plane
2x+y+z-7=0.
Putting the values of x, y and z in the equation of the plane, we get the value of a.
2( - \alpha + 3) + ( \alpha - 4) - (6 \alpha - 5) + 7 = 02(−α+3)+(α−4)−(6α−5)+7=0
= > 2 \alpha + 6 + \alpha - 4 + 6 \alpha - 5 - 7 = 0=>2α+6+α−4+6α−5−7=0
\begin{gathered}5 \alpha = 10 \\ \alpha = 2\end{gathered}
5α=10
α=2
Thus, the point of intersection is P(1, -2, 7).
Now, let P divide the line AB in the ratio m:
n.
By the section formula, we have
\begin{gathered}1 = \frac{2m + 3n}{m + n} \: \: \: \: \: \: \\ = > m + 2n = 0 \\ = > m = - 2n \: \: \: \: \\ = > \frac{m}{n} = > \frac{ - 2}{1} \end{gathered}
1=
m+n
2m+3n
=>m+2n=0
=>m=−2n
=>
n
m
=>
1
−2
Hence, P externally divides the line segment AB in the ratio 2:1
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