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Question
Find the coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)
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Solution 1
Equation of the line passing through (3, – 4, – 5) and (2, – 3, 1) is
⇒ (x - 1)(12) + (y - 2)6 + (z - 3)6 = 0
12x - 12 + 6y - 12 + 6z -18 = 0
12x + 6y + 6z - 42 = 0
2x + y + z - 7 = 0
2(-λ + 3) + 1(λ - 4) + (6λ - 5) - 7 = 0
- 2λ + 6 + λ - 4 + 6λ - 5 -7 = 0
5λ = 10 ⇒ λ = 2
∴ x = -2 +3, y = 2 - 4, z= 12 - 5
∴ x = 1, y = -2, z = 7
∴ Intersection point is (1,-2,7)
Solution 2
We know that the cartesian equation of a line passing through two points (x1, y1, z1)
and (x2, y2, z2)is given by
`(x - x_1)/(x_2 - x_1) = (y - y_1)/(y_2 - y_1) = (z - z_1)/(z_2 - z_1)`
So, the equation of a line passing through (3, –4, –5) and (2, –3, 1) is
`(x-3)/(2-3) = (y-(-4))/(-3-(-4)) = (z -(-5))/(1-(-5))`
`=> (x - 3)/(-1) = (y - (-4))/1 = (z-(-5))/6`
`=> (x - 3)/(-1) = (y+4)/1 = (z+5)/6`
Now, the coordinates of any point on this line are given by
`(x-3)/(-1) = (y +4)/1 = (z +5)/6 = k`
⇒ x = 3- k,y = k - 4, z = 6k - 5, where k is a constant
Let R(3 − k, k − 4, 6k − 5) be the required point of intersection.
Now,
Let the equation of a plane passing through (1, 2, 3) be
a(x − 1) + b(y − 2) + c(z − 3) = 0 .....(1)
Here, a, b, c are the direction ratios of the normal to the plane.
Since the plane (1) passes through (4, 2, −3), so
a(4 - 1) + b(2 - 2) + c(-3 - 3) = 0
⇒ 3a - 6c = 0 .....(2)
Also, the plane (1) passes through (0, 4, 3), so a(0 - 1) + b(4 - 2) + c(3 - 3) = 0
⇒ -a + 2b = 0 .....(3)
Solving (2) and (3) using the method of cross multiplication, we have
`a/(0 + 12) = b/(6 - 0) = c/(6 + 0)`
`=> a/12 = b/6 = c/6`
⇒ `a/2`= b = c = λ (Say)
⇒ a = 2λ, b = λ, c = λ
From (1), we get
2λ(x−1) + λ(y−2) + λ(z−3)=0
⇒ 2x + y +z −7=0 .....(4)
Putting x = 3−k, y = k−4, z = 6k−5 in (4), we get
⇒ 2(3 - k) + (k - 4) + (6k - 5) - 7 = 0
⇒5k − 10=0
⇒k = 2
Putting k = 2 in R(3 − k, k − 4, 6k − 5), we get
R(3 - k, k - 4, 6k - 5) = R(3 - 2, 2 - 4, 6 × 2 - 5) = R(1, -2,7)
Thus, the coordinates of the required point are (1, –2, 7) .
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