ID: Class: 12Subject: MathTopic: Vector AlgebraType: Very short (VSA)Year: 2025
Question:
If \(\vec{a}\) and \(\vec{b}\) are two non-collinear vectors, then find \(x\), such that \(\vec{\alpha}=(x-2)\vec{a}+\vec{b}\) and \(\vec{\beta}=(3+2x)\vec{a}-2\vec{b}\) are collinear.
Official Solution
Explanation:
If two vectors are collinear, one is a scalar multiple of the other.Let \(\vec{\alpha }=k\vec{\beta }\).
Substituting in \((x-2)\vec{a}+\vec{b}=k[(3+2x)\vec{a}-2\vec{b}]\)
\((x-2)\vec{a}+\vec{b}=k(3+2x)\vec{a}-2k\vec{b}\)
Since \(\vec{a}\) and \(\vec{b}\) are non-collinear, equating the coefficients of a and b both sides we get
\(x-2=k(3+2x)\) (Eq. 1) and
\(1=-2k\) (Eq. 2)
Solving for \(k\) and \(x\) we get \(k=-\frac{1}{2}\).
Substitute \(k\) into Eq. 1 we get
\(x-2=-\frac{1}{2}(3+2x)\)
\(2(x-2)=-(3+2x)\)
\(2x-4=-3-2x\)
\(4x=1\)
\(x=\frac{1}{4}\)
AI Teacher
Disclaimer: AI-generated content may contain errors. Please verify with standard textbooks.