If \( a , b , c \) are three non-zero..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> \( a , b , c... - Solved | MCA Prep" /> If \( a , b , c \) are three non-zero..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> \( a , b , c... - Solved | MCA Prep" /> If \( a , b , c \) are three non-zero..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 35

Mathematics Scalar and Vector Products Hard

<p>If <span class="math-tex">\(\rm \vec{a},\vec{b},\vec{c}\)</span><strong>&nbsp;</strong>are three non-zero vectors with no two of which are collinear, <span class="math-tex">\(\rm \vec{a}+2\vec{b}\)</span><strong>&nbsp;</strong>is collinear with <span class="math-tex">\(\rm \vec {c}\)</span>&nbsp;and <span class="math-tex">\(\rm \vec{b}+3\vec{c}\)</span>&nbsp;is collinear with <span class="math-tex">\(\rm \vec {a}\)</span>, then <strong><span class="math-tex">\(\rm |\vec{a} +2 \vec{b}+6\vec{c}|\)</span> </strong>will be equal to</p>

(A) Zero
(B) 9
(C) 1
(D) None of the above
✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Given that \( \vec a + 2 \vec b \) is collinear with \( \vec c \), we have: \[ \vec a + 2 \vec b = \lambda \vec c \quad \text{(1)} \] Also, \( \vec b + 3 \vec c \) is collinear with \( \vec a \), so: \[ \vec b + 3 \vec c = \mu \vec a \quad \text{(2)} \] Equating the value of \( 2 \vec b \) from the above two equations, we get: \[ \lambda \vec c - \vec a = 2 \mu \vec a - 6 \vec c \] By comparison, we find that \( \lambda = -6 \). Substituting this value into the first equation, we get: \[ \vec a + 2 \vec b = -6 \vec c \] Or: \[ \left| \vec a + 2 \vec b + 6 \vec c \right| = 0 \]

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