1
Question
→r1=^i−^j+3^k+λ(^i−^j+^k)
→r2=2^i+4^j+6^k+μ(2^i+^j+3^k)
→r2=2^i+4^j+6^k+μ(2^i+^j+3^k)
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Solution
In this case the two sets of direction ratios are 1:−1:1 and 2:1:3.
They are not equal, so these two lines are not parallel.
Now, if the lines intersect it will be at a point where r1=r2 i.e., when
(1+α)i−(1+α)j+(3+α)k=2(1+μ)i+(4+μ)j+(6+3μ)k
Equating the coefficients of i and j, we have
1+α=2(1+μ) and −(1+α)=4+μ
Hence, μ=−2 and α=−3
With these values of α and μ, the coefficients of k become
for first line 3+α=0
for second line 6+3μ=0
Both have equal values.
So r1=r2 when α=−3,μ=−2
Therefore, the lines intersect at the point with position vector.
(1−3)i−(1−3)j+(3−3)k i.e., −2i+2j
They are not equal, so these two lines are not parallel.
Now, if the lines intersect it will be at a point where r1=r2 i.e., when
(1+α)i−(1+α)j+(3+α)k=2(1+μ)i+(4+μ)j+(6+3μ)k
Equating the coefficients of i and j, we have
1+α=2(1+μ) and −(1+α)=4+μ
Hence, μ=−2 and α=−3
With these values of α and μ, the coefficients of k become
for first line 3+α=0
for second line 6+3μ=0
Both have equal values.
So r1=r2 when α=−3,μ=−2
Therefore, the lines intersect at the point with position vector.
(1−3)i−(1−3)j+(3−3)k i.e., −2i+2j
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