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ScanSolve AI Answer
The two lines L_1 and L_2 are given by: L_1: r = i - 2j + 3k + (2i - j + ak) L_2: r = -i - j - k + (3i - 2j - 2k) The direction vector of L_1 is d_1 = 2 \\ -1 \\ a . The direction vector of L_2 is d_2 = 3 \\ -2 \\ -2 . a) The value of the constant a. Step 1: Use the condition that L_1 and L_2 are perpendicular. If two lines are perpendicular, their direction vectors are orthogonal, meaning their dot product is zero: d_1 · d_2 = 0. (2)(3) + (-1)(-2) + (a)(-2) = 0 6 + 2 - 2a = 0 8 - 2a = 0 Step 2: Solve for a. 2a = 8 a = 4 b) The position vector of the point of intersection of L_1 and L_2. Step 1: Equate the position vectors of the two lines. Substitute a=4 into the equation for L_1: i - 2j + 3k + (2i - j + 4k) = -i - j - k + (3i - 2j - 2k) Equate the components: For i: 1 + 2 = -1 + 3 2 - 3 = -2 (1) For j: -2 - = -1 - 2 - + 2 = 1 (2) For k: 3 + 4 = -1 - 2 4 + 2 = -4 2 + = -2 (3) Step 2: Solve the system of linear equations for and . From equation (2), we can express in terms of : = 2 - 1 Substitute this into equation (3): 2(2 - 1) + = -2 4 - 2 + = -2 5 = 0 = 0 Now substitute =0 back into the expression for : = 2(0) - 1 = -1 Verify these values with equation (1): 2(-1) - 3(0) = -2 - 0 = -2. The values are consistent. Step 3: Substitute the value of (or ) into the respective line equation to find the intersection point. Using L_1 with = -1: r = i - 2j + 3k + (-1)(2i - j + 4k) r = i - 2j + 3k - 2i + j - 4k r = (1-2)i + (-2+1)j + (3-4)k r = -i - j - k c) The Cartesian equation of the plane containing L_1 and L_2. Step 1: Find a normal vector to the plane. The plane contains both lines, so its normal vector n must be perpendicular to both direction vectors d_1 and d_2. We can find n by taking the cross product of d_1 and d_2. Using a=4, d_1 = 2 \\ -1 \\ 4 and d_2 = 3 \\ -2 \\ -2 . n = d_1 × d_2 = i & j & k \\ 2 & -1 & 4 \\ 3 & -2 & -2 n = i((-1)(-2) - (4)(-2)) - j((2)(-2) - (4)(3)) + k((2)(-2) - (-1)(3)) n = i(2 - (-8)) - j(-4 - 12) + k(-4 - (-3)) n = 10i + 16j - k Step 2: Use the normal vector and a point on the plane to find the Cartesian equation. The Cartesian equation of a plane is Ax + By + Cz = D, where A \\ B \\ C is the normal vector. So, the equation is 10x + 16y - z = D. We know the point of intersection P(-1, -1, -1) lies on the plane. Substitute these coordinates to find D: 10(-1) + 16(-1) - (-1) = D -10 - 16 + 1 = D -25 = D Step 3: Write the Cartesian equation of the plane. 10x + 16y - z = -25 What's next?