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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
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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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