Let a line passing through the point (4,1,0) intersect the line L1:2x−1=3y−2=4z−3 at the point A(α,β,γ) and the line L2:x−6=y−z+4 at the point B(a,b,c). Then det1αa0βb1γc is equal to:
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Step-by-Step Solution
Step 1: Represent points on lines L1 and L2
First, we represent the general points on lines L1 and L2. For L1, we set 2x−1=3y−2=4z−3=λ1. This gives us the coordinates of point A in terms of λ1. For L2, we have x−6=y and y=z−4. Let y=λ2. Then x=λ2+6 and z=λ2+4. So, the coordinates of point B are (λ2+6,λ2,λ2+4).
Step 2: Use collinearity of P, A, B to find parameters
Since points P(4,1,0), A(α,β,γ), and B(a,b,c) are collinear, the direction ratios of PA and PB must be proportional. We can write this as a−4α−4=b−1β−1=c−0γ−0. Substituting the expressions for A and B in terms of λ1 and λ2 into this proportionality, we get a system of equations.
Step 3: Solve for parameters λ1 and λ2
From the proportionality of direction ratios, we get two equations:
(2λ1−3)(3λ1+2−λ2)=(3λ1+1)(2λ1−λ2−5)
and (3 λ1+1 )(4 λ1−λ2−1 )=(4 λ1+3 )(3 λ1+2−λ2). Solving these simultaneous equations yields λ1=−1 and λ2=3.
Step 4: Find coordinates of A and B
Now that we have the values of λ1 and λ2, we can substitute them back into the expressions for points A and B. For A, using λ1=−1, we get A=(2(−1)+1,3(−1)+2,4(−1)+3)=(−1,−1,−1). For B, using λ2=3, we get B=(3+6,3,3+4)=(9,3,7).
Step 5: Calculate the determinant
Finally, we need to calculate the determinant of the given matrix using the coordinates of A(α,β,γ)=(−1,−1,−1) and B(a,b,c)=(9,3,7). The determinant is 1((−1)(7)−(−1)(3))−0((−1)(7)−(−1)(9))+1((−1)(3)−(−1)(9)). This simplifies to 1(−7+3)+1(−3+9)=1(−4)+1(6)=−4+6=2.