Let \( a , b \) and \( c \) be three non-zero vectors, no two of which are collinear. If the vector \( a +2 b \) is collinear with \( c \) and \( b +3 c \) is collinear with <span...

\[ Let a+2b = xc and b+3c = ya. \] \[ Then a+2b+6c = (x+6)c \] \[ and also, a+2b+6c = (1+2y)a. \] \[ So (x+6)c = (1+2y)a. \] \[ Since a, b, and c are non-zero and non-collinear, \] \[ we have x+6 = 0 \] \[ and 1+2y = 0, i.e. x=-6 and y=-1/2. \] \[ In either case, we have a+2b+6c = 0. \]

Let <span class=

Question 24

Mathematics Scalar and Vector Products Hard

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors, no two of which are collinear. If the vector \(\vec{a}+2\vec{b}\) is collinear with \(\vec{c}\) and \(\vec{b}+3\vec{c}\) is collinear with \(\vec{a}\), then \(\vec{a} + 2\vec{b}+6\vec{c}\) is equal to

(A) \(\lambda \vec{a}\)

(B) \(\lambda \vec{b}\)

(C) \(\lambda \vec{c}\)

(D) \(\vec{0}\)

✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ \text{Let } a+2b = xc \quad \text{and} \quad b+3c = ya. \] \[ \text{Then } a+2b+6c = (x+6)c \] \[ \text{and also, } a+2b+6c = (1+2y)a. \] \[ \text{So } (x+6)c = (1+2y)a. \] \[ \text{Since } a, b, \text{ and } c \text{ are non-zero and non-collinear,} \] \[ \text{we have } x+6 = 0 \] \[ \text{and } 1+2y = 0, \text{ i.e. } x=-6 \text{ and } y=-\frac{1}{2}. \] \[ \text{In either case, we have } a+2b+6c = 0. \]

Full-Length Test Series

Prepare for NIMCET Under Real Exam Conditions

Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.

Start Free Mock Test Series →

Question 24

📘 Step-by-Step Solution & Detailed Explanation:

Practice Similar Scalar and Vector Products Questions

Prepare for NIMCET Under Real Exam Conditions