<p>If \(\vec{a} = \vec{i} + 6\vec{j} + 3\vec{k}\); \(\vec{b} = 3\vec{i} + 2\vec{j} + \vec{k}\) and \(\vec{c} = (a+1)\vec{i} + (b-1)\vec{j} + \vec{k}\) are linearly dependent vectors and \(|\vec{c}| = \sqrt{6}\), then the possible value(s) of \((a + b)\) can be:</p>
Step-by-Step Solution
Key Concept: If three vectors are linearly dependent, one vector must be expressible as a linear combination of the others. Here, since we have only two independent vectors (a and b), vector c must be a scalar multiple of their linear combination, which gives us constraints on a and b. Combined with the magnitude condition |c| = √6, we can solve for specific values.
Step 1: Set up the linear dependence condition. Since vectors a, b, and c are linearly dependent, we can write: c = λa + μb for some scalars λ, μ (a+1)i + (b-1)j + k = λ(i + 6j + 3k) + μ(3i + 2j + k) (a+1)i + (b-1)j + k = (λ + 3μ)i + (6λ + 2μ)j + (3λ + μ)k Step 2: Equate coefficients. From the i-component: a + 1 = λ + 3μ ... (1) From the j-component: b - 1 = 6λ + 2μ ... (2) From the k-component: 1 = 3λ + μ ... (3) Step 3: Solve equations (1), (2), (3) using equation (3). From (3): μ = 1 - 3λ Substitute in (1): a + 1 = λ + 3(1 - 3λ) = λ + 3 - 9λ = 3 - 8λ Therefore: a = 2 - 8λ ... (4) Substitute in (2): b - 1 = 6λ + 2(1 - 3λ) = 6λ + 2 - 6λ = 2 Therefore: b = 3 ... (5) Step 4: Use the magnitude condition |c| = √6. |c|^2 = (a+1)^2 + (b-1)^2 + 1^2 = 6 With b = 3: (a+1)^2 + (3-1)^2 + 1 = 6 (a+1)^2 + 4 + 1 = 6 (a+1)^2 = 1 a + 1 = ±1 Therefore: a = 0 or a = -2 Step 5: Find possible values of (a + b). Since b = 3: If a = 0: a + b = 0 + 3 = 3 If a = -2: a + b = -2 + 3 = 1 Step 6: Verify both solutions satisfy linear dependence. For a = 0, b = 3: From equation (4), λ = 1/4. Check: μ = 1 - 3(1/4) = 1/4. Verify all three equations: ✓ For a = -2, b = 3: From equation (4), λ = 1/2. Check: μ = 1 - 3(1/2) = -1/2. Verify all three equations: ✓ ∴ Answer: The possible values of (a + b) are 1 and 3, which corresponds to options a,c
Correct Answer: a,c