0
$\begingroup$

Plane $P$ contains the lines

\begin{align}L_1:&\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{1}=3\\ L_2:&\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}\end{align}

Prove that $L_1$ and $L_2$ intersect and the equation of plane is $x-y+1=0$.

I got the answer the answer of first part by susbtituting parametric equation of points in other equation of line .

But I am stuck how to find the plane

|cite|improve this question
$\endgroup$
  • $\begingroup$ Is line L1 correct? I'm talking about z part. According to me it might be '/' instead of '=' $\endgroup$ – sgrmshrsm7 Apr 21 '17 at 10:49
  • 1
    $\begingroup$ What do you mean on $L_1$ having $=3$? $\endgroup$ – Juniven Apr 21 '17 at 11:23
1
$\begingroup$

\begin{align}L_1:\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}=\lambda\\ L_2:\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}=\epsilon\end{align}

Vector parallel to $L_1:\vec{r_1}=k\underline{i}+2 \underline{j}+3\underline{k}$

Vector parallel to $L_2:\vec{r_2}=2\underline{i}+k\underline{j}+3\underline{k}$

Normal vector to required plane: $\vec{n}=\vec{r_1}\times \vec{r_2}$

Point on plane will be $(1,2,3)$ as it lies on line $L_1$

So equation of required plane is : $$(\vec{r}-\underline{i}-2\underline{j}-3\underline{k})\cdot \vec{n}=0$$

Where $\vec{r}=x\underline{i}+y\underline{j}+3\underline{k}$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I don't know how to write j cap and k cap if anybody knows then he can Edit the answer. I can write î bcz there is option on my keyboard. Thank You!! $\endgroup$ – sgrmshrsm7 Apr 21 '17 at 11:15
1
$\begingroup$

As you already did first part I'll tell about second part. $\frac{x-2}{2}=\frac{y-3}{k}=\frac{z-4}{3}=a$ where a is any variable. Since you can find value of k yourself so just subtitue it in the above equation I wrote. Now you can get point let name it $P_2$ $(2a+2, ka+3, 3a+4)$ and from $L_1$ you can get $P_1$ the find direction ratios and now you're ready to form equation of plane.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ are you sure that $L_1,L_2$ intersect each other? $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 11:08
  • $\begingroup$ @Dr.SonnhardGraubner the person already said he did first part that's why I didn't do it. $\endgroup$ – Iti Shree Apr 21 '17 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.