Question

Give an example of three nonzero vectors u, v and w in ${\mathrm{ℝ}}^3$ such that $u\times v=u\times w$ but $v\ne w$. What geometric relationship must the three vectors have for this to happen?

Short answer

Let $u=(1,0,0),v=(2,1,1)\mathrm{and}w=(4,1,1)$.

If $u\times v=u\times w$, then u is parallel to $v-w$.

Step-by-step solution

Step 1. Given Information 

Give an example of three nonzero vectors u, v and w in ${\mathrm{ℝ}}^3$ such that role="math" localid="1649322061069" $u\times v=u\times w$ but $v\ne w$. What geometric relationship must the three vectors have for this to happen?

Step 2. Let u=(1,0,0), v=(2,1,1) and w=(4,1,1)

Now finding the value of $u\times v$.

$$\begin{gathered} u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 0& 0\\ 2& 1& 1\end{array}\right] \\ u\times v=\left(\right(0\left)\right(1)-(1\left)\right(0\left)\right)i+\left(\right(1\left)\right(1)-(2\left)\right(0\left)\right)j+\left(\right(1\left)\right(1)-(2\left)\right(0\left)\right)k \\ u\times v=(0+0)i+(1-0)j+(1-0)k \\ u\times v=0i+1j+1k \end{gathered}$$

Step 3. Now finding the value of

$$\begin{gathered} u\times w=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 0& 0\\ 4& 1& 1\end{array}\right] \\ u\times w=\left(\right(0\left)\right(1)-(1\left)\right(0\left)\right)i+\left(\right(1\left)\right(1)-(4\left)\right(0\left)\right)j+\left(\right(1\left)\right(1)-(4\left)\right(0\left)\right)k \\ u\times w=(0+0)i+(1-0)j+(1-0)k \\ u\times w=0i+1j+1k \end{gathered}$$

Hence,$u\times v=u\times w=0i+1j+1k$but$v\ne w$.

Step 4. Now finding the relation of three vectors.

If $u\times v=u\times w$, then u is parallel to $v-w$.