# Show that there exists a point $D$ on the line $AB$ whose position vector is $\vec{d}=2t\hat{i}+(1-2t)\hat{j}+(t-4)\hat{k}$

If $A,B,C$ are three vertices of a triangle $ABC$ with position vectors $\vec{a},\vec{b},\vec{c}$ respectively,such that $\vec{a}=\frac{1}{2}(2\hat{i}-\vec{r}-7\hat{k});\vec{b}=(3\vec{r}+\vec{j}-4\hat{k});\vec{c}=(22\hat{i}-11\hat{j}-9\vec{r})$.A vector $\vec{p}=2\hat{j}-\hat{k}$ is such that $\vec{r}+\vec{p}$ is parallel to $\hat{i}$ and $(\vec{r}-2\hat{i})$ is parallel to $\vec{p}$.Show that there exists a point $D$ on the line $AB$ whose position vector is $\vec{d}=2t\hat{i}+(1-2t)\hat{j}+(t-4)\hat{k}$.

Since $\vec{r}+\vec{p}$ is parallel to $\hat{i}$.So their cross product is zero.
$(\vec{r}+\vec{p})\times\hat{i}=0$
$(\vec{r}+2\hat{j}-\hat{k})\times\hat{i}=0$
$\vec{r}\times\hat{i}=\hat{j}+2\hat{k}......................(1)$
Since $\vec{r}-2\hat{i}$ is parallel to $\vec{p}$.So their cross product is zero.
$(\vec{r}-2\hat{i})\times\vec{p}=0\Rightarrow(\vec{r}-2\hat{i})\times(2\hat{j}-\hat{k})=0$
$2\vec{r}\times\hat{j}-\vec{r}\times\hat{k}-4\hat{k}-2\hat{j}=0..............(2)$

• what role does the point $C$ play? – David Holden Nov 14 '15 at 15:46
• $C$ plays the role in the second part of the question which i did not type here. – diya Nov 14 '15 at 15:48
• ok, well have your tried setting $\vec{r}=r_1\vec{i}+r_2\vec{j}+r_3\vec{k}$ and using the cross-products you mention to compute the components $r_j$? once you have those you can find $\vec{a}$ and $\vec{b}$ and then try solving $t\vec{a}+(1-t)\vec{b}=\vec{d}$ – David Holden Nov 14 '15 at 15:55
• What is the source of your question? Is it from a book or an in-class assignment? – Tim Thayer Nov 14 '15 at 16:13
• What is the title of your book? I'm always looking for good sources of challenging problems... – Tim Thayer Nov 14 '15 at 16:16

There are two common ways to show vectors are parallel. You tried that the cross product of the two vectors is zero. In problems like this, usually it's better to use that two vectors are parallel iff one vector is a scalar multiple of the other.

Finding $\vec{r}$

We have that $\vec{r}+\vec{p}$ is parallel to $\vec{\imath}$, so for some scalar $m$, we have that $\vec{r}+\vec{p}=m\vec{\imath}$. Using $\vec{p}=2\vec{\jmath}-\vec{k}$ and solving for $\vec{r}$ yields $\vec{r}=m\vec{\imath}-2\vec{\jmath}+\vec{k}$.

We also have that $\vec{r}-2\vec{\imath}$ is parallel to $\vec{p}$, so for some scalar $n$, we have that $\vec{r}-2\vec{\imath}=n\vec{p}$. Using $\vec{p}=2\vec{\jmath}-\vec{k}$ and solving for $\vec{r}$ yields $\vec{r}=2\vec{\imath}+2n\vec{\jmath}-n\vec{k}$.

Comparing the two expressions for $\vec{r}$, we find that $m=2$ and $n=-1$. Thus, we have that $\vec{r}=2\vec{\imath}-2\vec{\jmath}+\vec{k}$.

Finding the equation of $AB$

Knowing $\vec{r}$, we find $\vec{a}$ and $\vec{b}$ explicitly.

\begin{align*} \vec{a}&=\frac{1}{2}\left(2\vec{\imath}-\vec{r}-7\vec{k}\right)\\ &=\frac{1}{2}\left(2\vec{\imath}-(2\vec{\imath}-2\vec{\jmath}+\vec{k})-7\vec{k}\right)\\ &=\vec{\jmath}-4\vec{k} \end{align*}

\begin{align*} \vec{b}&=3\vec{r}+\vec{\jmath}-4\vec{k}\\ &=3(2\vec{\imath}-2\vec{\jmath}+\vec{k})+\vec{\jmath}-4\vec{k}\\ &=6\vec{\imath}-5\vec{\jmath}-\vec{k} \end{align*}

A vector in the direction of line $AB$ is $\vec{b}-\vec{a}=6\vec{\imath}-6\vec{\jmath}+3\vec{k}$, or more simply, $2\vec{\imath}-2\vec{\jmath}+\vec{k}$.

Taking the direction vector along with point A yields the vector equation of line AB: \begin{align*} \vec{d}&=\vec{\jmath}-4\vec{k}+t\left(2\vec{\imath}-2\vec{\jmath}+\vec{k}\right)\\ &=2t\vec{\imath}+(1-2t)\vec{\jmath}+(t-4)\vec{k} \end{align*}

And now, we are done.