Q.Show that the vectors $-\hat{i}-2\hat{j}-6\hat{k}$, $2\hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+5\hat{k}$ form a right angled triangle.

For non-zero vectors $\vec a,\vec b$ with included angle $\theta$ ($0\le\theta\le\pi$), the scalar (dot) product is the number $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta.$

Geometric meaning (projection). $\vec a\cdot\vec b=|\vec a|\times(\text{projection of }\vec b\text{ on }\vec a)$, and the projection of $\vec b$ on $\vec a$ is $\dfrac{\vec a\cdot\vec b}{|\vec a|}$ (symmetrically, projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$).

Core properties.

• Commutative: $\vec a\cdot\vec b=\vec b\cdot\vec a$.
• Sign follows the angle: positive for $0\le\theta<\pi/2$, zero at $\theta=\pi/2$, negative for $\pi/2<\theta\le\pi$. In particular $\vec a\cdot\vec b=0 \iff \vec a=\vec 0$ or $\vec b=\vec 0$ or $\vec a\perp\vec b$ — for two non-zero vectors, $\vec a\cdot\vec b=0$ is exactly the perpendicularity test.
• $\vec a\cdot\vec a=|\vec a|^2$ (often written $a^2$), so $|\vec a|=\sqrt{\vec a\cdot\vec a}$.
• $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ and $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$ (they're mutually perpendicular unit vectors).
• Distributive: $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$, and likewise for subtraction and for the right factor. …