A line with direction cosines proportional to `1,-5, a n d-2` meets lines `x=y+5=z+1a n dx+5=3y=2zdot` The coordinates of each of the points of the in

Question

A line with direction cosines proportional to `1,-5, a n d-2` meets lines `x=y+5=z+1a n dx+5=3y=2zdot` The coordinates of each of the points of the in

A line with direction cosines proportional to `1,-5, a n d-2` meets lines `x=y+5=z+1a n dx+5=3y=2zdot` The coordinates of each of the points of the intersection are given by a. `(2,-3,1)` b. `(1,2,3)` c. `(0,5//3,5//2)` d. `(3,-2,2)`
A. (2,-3,1)
B. (1,2,3)
C. `(0,5//3,5//2)`
D. (3,-2,2)

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - a, b
Let the coordinates of the point(s) be a, b and c. Therefore, the equation of the line passing through `(a, b, c)` and whose direction ratios are `1, -5 and -2` is
`" "(x-a)/(1)=(y-b)/(-5)= (z-c)/(-2)" "`(i)
Line (i) intersects the line,
`" "(x)/(1)=(y+5)/(1)= (z+1)/(1)" "` (ii)
Therefore, these are coplanar.
`" "|{:(1,,-5,,-2),(1,,1,,1),(a,,b+5,,c+1):}|=0`
or `" "a+b-2c+3=0`
Also, by using same procedure with the second equation, we get the condition
`" "11a+15b-32c+55=0`