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Question
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes
Solution
The straight line passes through the points (6, 7, 4) and (8, 4, 9)
∴ Direction ratios are 2, – 3, 5
So the straight line is parallel to `2hat"i" - 3hat"j" + 5hat"k"`
∴ Vector equation is
`vec"r" = (6hat"i" + 7hat"j" + 4hat"k") + "t"(2hat"i" - 3hat"j" + 5hat"k")`
or
`vec"r" = (8hat"i" + 4hat"j" + 9hat"k") + "t"(2hat"i" - 3hat"j" + 5hat"k")`, t ∈ R
Cartesian equation
`(x - 6)/2 = (y - 7)/(- 3) = (z - 4)/5`
or
`(x - 8)/2 = (y - 4)/(- 3) = (z - 9)/5`
`(x - 6)/2 = (y - 7)/(-3) = (z - 4)/5` = t
`(x - 8)/2 = (y - 4)/(-3) = (z - 9)/5` = s
An arbitrary point on the straight line is of the form
(2t + 6, – 3t + 7, 5t + 4)
or
(2s + 8, – 3s + 4, 5s + 9)
(i) xz plane mean y = 0
– 3t + 7 = 0
3t = 7
t = `7/3`
Point = `(2(7/3) + 6, 0, 5(7/3) + 4)`
= `(32/3, 0, 47/3)`
(ii) The straight Line cuts yz-plane
So we get x = 0
2t + 6 = 0
⇒ 2t = – 6
t = – 3
– 3t + 7 = – 3(– 3) + 7
= 9 + 7
= 16
5t + 4 = 5(– 3) + 4
= – 15 + 4
= – 11
The required point (0, 16, – 11).
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