Q.Using vectors, show that the points $A(1,2,7)$, $B(2,6,3)$ and $C(3,10,-1)$ are collinear.

For parallel vectors, corresponding components must be in the same ratio; use the $\hat i$ or $\hat j$ components to find the scale factor, then apply it to $\hat k$.

Given $\vec a=\hat i+3\hat j-\hat k$, $\vec b=2\hat i+6\hat j+\lambda\hat k$, with $\vec a\parallel\vec b$, so $\vec b=k\vec a$ for some scalar $k$.

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Setting $|m\vec a|=1$ and using $|m\vec a|=|m||\vec a|$ shows $m\vec a$ is a unit vector exactly when $|\vec a|=1/|m|$.

1. Let $a=|\vec a|$ denote the magnitude (length) of the non-zero vector $\vec a$.
2. For a scalar $m$, the magnitude of $m\vec a$ is $|m\vec a|=|m|\cdot|\vec a|=|m|\,a$.
3. $m\vec a$ is a unit vector precisely when its magnitude equals $1$: $|m|\,a=1$. …

Since a unit vector has magnitude $1$, $|m\vec a|=1$ directly forces $|\vec a|=1/|m|$.

1. $m\vec a$ is a unit vector precisely when its magnitude equals $1$: $|m\vec a|=1$.
2. Magnitude of a scalar multiple: $|m\vec a|=|m|\,|\vec a|$.
3. So the condition becomes $|m|\,|\vec a|=1$.
4. Solving for $|\vec a|$ (denoted $a$ in the options, since $m\ne0$): $|\vec a|=\dfrac{1}{|m|}$, i.e. $a=\dfrac1{|m|}$. …