Multiplying Vectors — Question 4

Question 4

Find a parameter that makes two vectors perpendicular

Dimensionless vectors are $\vec a=(t,2,-1)$ and $\vec b=(3,-1,4)$, where t is real.

1. Find t for which the vectors are perpendicular.

2. Can any t make these two vectors parallel? Use their component ratios.

3. Does $\vec a\cdot\vec b=0$ always mean a $90^\circ$ angle for arbitrary vectors? State the needed qualification.

Question 4 - Solution

Find a parameter that makes two vectors perpendicular

(a) Set the scalar product to zero.

$$\vec a\cdot\vec b=3t-2-4=0\Rightarrow\boxed{t=2},\quad a=3,\quad b=\sqrt{26}$$ $\vec a\cdot\vec b=3t-2-4=3t-6$. Thus $t=2$. At this value, $|\vec a|=3$ and $|\vec b|=\sqrt{26}$, so both are nonzero and $\cos\theta=0$ genuinely gives $\theta=90^\circ$.

(b) Check whether one common multiplier exists.

$$\vec a=\lambda\vec b\Rightarrow\lambda=2/(-1)=-2\quad\text{and}\quad\lambda=(-1)/4=-1/4\quad\text{incompatible}$$ Parallelism would require $\vec a=\lambda\vec b$. The y components demand $\lambda=-2$, whereas the z components demand $\lambda=-1/4$. They disagree regardless of t, so no t makes the vectors parallel.

(c) Exclude undefined directions.

$$\cos\theta=\frac{\vec a\cdot\vec b}{ab}\quad(a,b>0);\qquad\vec0\cdot\vec b=0\text{ but angle undefined}$$ The zero vector has zero dot product with every vector, yet has no direction and therefore no defined angle with another vector. The inference of a right angle requires both vectors to be nonzero. In this problem the fixed y and z entries already guarantee that condition.