<p>Let <strong>a</strong> = <strong>i</strong> + <strong>j</strong> and <strong>b</strong> = 2<strong>i</strong> - <strong>k</strong>, then the point of intersection of the lines <strong>r</strong> × <strong>a</strong> = <strong>b</strong> × <strong>a</strong> and <strong>r</strong> × <strong>b</strong> = <strong>a</strong> × <strong>b</strong> is</p>
Step-by-Step Solution
Key Concept: To find the intersection point of two lines given in the form r × a = b × a and r × b = a × b, we need to find a position vector r that satisfies both equations simultaneously. The key is to use the vector triple product formula and properties of cross products.
Step 1: Given vectors are a = i + j and b = 2i - k. We need to solve the system: r × a = b × a ... (1) r × b = a × b ... (2) Step 2: From equation (1): r × a = b × a, we can write (r - b) × a = 0. This means (r - b) is parallel to a, so r - b = λa for some scalar λ. Therefore: r = b + λa = (2i - k) + λ(i + j) = (2 + λ)i + λj - k ... (i) Step 3: From equation (2): r × b = a × b, we can write (r - a) × b = 0. This means (r - a) is parallel to b, so r - a = μb for some scalar μ. Therefore: r = a + μb = (i + j) + μ(2i - k) = (1 + 2μ)i + j - μk ... (ii) Step 4: Comparing equations (i) and (ii) component-wise: From i-component: 2 + λ = 1 + 2μ → λ - 2μ = -1 ... (A) From j-component: λ = 1 ... (B) From k-component: -1 = -μ → μ = 1 ... (C) Step 5: From (B), λ = 1. Checking with (A): 1 - 2(1) = -1 ✓ and (C) gives μ = 1 ✓ Step 6: Substituting λ = 1 into equation (i): r = (2 + 1)i + (1)j - k = 3i + j - k Step 7: Verification using μ = 1 in equation (ii): r = (1 + 2(1))i + j - (1)k = 3i + j - k ✓ Step 8: The point of intersection is (3, 1, -1), which corresponds to option (b). ∴ Answer: The intersection point is (3, 1, -1)
Correct Answer: Unknown