Vector Algebra
Collinearity of Vectors
Grade None
Question:
<p>Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of these are collinear. If the vector \(\vec{a} + 2\vec{b}\) is collinear with \(\vec{c}\) and \(\vec{b} + 3\vec{c}\) is collinear with \(\vec{a}\) (\(\lambda\) being some non-zero scalar) then \(\vec{a} + 2\vec{b} + 6\vec{c}\) equals</p>
<p>\(\lambda \vec{a}\)</p>
<p>\(\lambda \vec{b}\)</p>
<p>\(\lambda \vec{c}\)</p>
<p>\(0\)</p>
Step-by-Step Solution
Key Concept: When two vectors are collinear, one is a scalar multiple of the other. Use this condition to create a system of equations relating a, b, and c, then solve for the coefficients to express a + 2b + 6c as a scalar multiple of one vector.
Step 1: From the given conditions, use collinearity:
Since (a + 2b) is collinear with c: a + 2b = μc for some scalar μ
Since (b + 3c) is collinear with a: b + 3c = λa for some scalar λ Step 2: Express b from the second equation:
b + 3c = λa → b = λa - 3c Step 3: Substitute into the first equation:
a + 2(λa - 3c) = μc
a + 2λa - 6c = μc
a(1 + 2λ) = μc + 6c
a(1 + 2λ) = (μ + 6)c Step 4: Since a and c are not collinear (given), this is only possible if:
1 + 2λ = 0 → λ = -1/2
μ + 6 = 0 → μ = -6 Step 5: Now find a + 2b + 6c using a + 2b = μc = -6c:
a + 2b + 6c = -6c + 6c = 0 ∴ Answer: D
Correct Answer: D