1
Question
Let L1, and L2 denotes the lines
→r=^i+λ(−^i+2^j+2^k),λ∈R and
→r=μ(2^i−^j+2^k),μ∈R respectively.
If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following option describe(s) L3?
Let L1, and L2 denotes the lines
→r=^i+λ(−^i+2^j+2^k),λ∈R and
→r=μ(2^i−^j+2^k),μ∈R respectively.
If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following option describe(s) L3?
Open in App
Solution
The correct option is C →r=29(4^i+^j+^k)+t(2^i+2^j−^k),t∈R
Equation of line L1:→r=^i+λ(−^i+2^j+2^k), λ∈R
Equation of line L2:→r=μ(2^i−^j+2^k), μ∈R
Therefore, Point on lines L1 and L2 are A(1−λ,2λ,2λ) and B(2μ,−μ,2μ)
∴−−→AB=(2μ+λ−1)^i+(−μ−2λ)^j+(2μ−2λ)^k
L3 is perpendicular to both L1 and L2
∴L3∥(L1×L2)
L3=m∣∣
∣
∣∣^i^j^k−1222−12∣∣
∣
∣∣=3m(2^i+2^j−^k)=t(2^i+2^j−^k m,t∈R)
Direction ratios of AB will be (2t,2t,−t) which is propotional to (2,2,−1)
Hence, 2μ+λ−12=−μ−2λ2=2μ−2λ−1=c (Let)
2μ+λ−1=2c...(i)
−μ−2λ=2c...(ii)
2μ−2λ=−c...(iii)
On solving equations we get,
λ=19, μ=29
∴A=(89,29,29), B=(49,−29,49)
Mid Point of −−→AB=(23,0,13)
∴ Equation of line L3 can be
→r=89^i+29^j+29^k+t(2^i+2^j−^k) t∈R
→r=29(4^i+^j+^k)+t(2^i+2^j−^k) t∈R or
→r=49^i−29^j+29^k+t(2^i+2^j−^k) t∈R
→r=29(2^i−^j+^k)+t(2^i+2^j−^k) t∈Ror
→r=23^i+13^k+t(2^i+2^j−^k) t∈R
→r=13(2^i+^k)+t(2^i+2^j−^k) t∈R
Equation of line L1:→r=^i+λ(−^i+2^j+2^k), λ∈R
Equation of line L2:→r=μ(2^i−^j+2^k), μ∈R
Therefore, Point on lines L1 and L2 are A(1−λ,2λ,2λ) and B(2μ,−μ,2μ)
∴−−→AB=(2μ+λ−1)^i+(−μ−2λ)^j+(2μ−2λ)^k
L3 is perpendicular to both L1 and L2
∴L3∥(L1×L2)
L3=m∣∣ ∣ ∣∣^i^j^k−1222−12∣∣ ∣ ∣∣=3m(2^i+2^j−^k)=t(2^i+2^j−^k m,t∈R)
Direction ratios of AB will be (2t,2t,−t) which is propotional to (2,2,−1)
Hence, 2μ+λ−12=−μ−2λ2=2μ−2λ−1=c (Let)
2μ+λ−1=2c...(i)
−μ−2λ=2c...(ii)
2μ−2λ=−c...(iii)
On solving equations we get,
λ=19, μ=29
∴A=(89,29,29), B=(49,−29,49)
Mid Point of −−→AB=(23,0,13)
∴ Equation of line L3 can be
→r=89^i+29^j+29^k+t(2^i+2^j−^k) t∈R
→r=29(4^i+^j+^k)+t(2^i+2^j−^k) t∈R or
→r=49^i−29^j+29^k+t(2^i+2^j−^k) t∈R
→r=29(2^i−^j+^k)+t(2^i+2^j−^k) t∈Ror
→r=23^i+13^k+t(2^i+2^j−^k) t∈R
→r=13(2^i+^k)+t(2^i+2^j−^k) t∈R
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
