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Exercise 6.4 · Q57

Q.Show that lines r⃗=(2j^−3k^)+λ(i^+2j^+3k^)\vec{r}=(2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k}) and r⃗=(2i^+6j^+3k^)+μ(2i^+3j^+4k^)\vec{r}=(2\hat{i}+6\hat{j}+3\hat{k})+\mu(2\hat{i}+3\hat{j}+4\hat{k}) are coplanar. Find the equation of the plane determined by them.

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Here a⃗1=2j^−3k^=(0,2,−3)\vec a_1=2\hat j-3\hat k=(0,2,-3), b⃗1=i^+2j^+3k^=(1,2,3)\vec b_1=\hat i+2\hat j+3\hat k=(1,2,3), a⃗2=2i^+6j^+3k^=(2,6,3)\vec a_2=2\hat i+6\hat j+3\hat k=(2,6,3), b⃗2=2i^+3j^+4k^=(2,3,4)\vec b_2=2\hat i+3\hat j+4\hat k=(2,3,4).

a⃗2−a⃗1=(2,4,6)\vec a_2-\vec a_1=(2,4,6).

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=∣246123234∣=2(8−9)−4(4−6)+6(3−4)=2(−1)−4(−2)+6(−1)=−2+8−6=0.(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=\begin{vmatrix}2&4&6\\1&2&3\\2&3&4\end{vmatrix}=2(8-9)-4(4-6)+6(3-4)=2(-1)-4(-2)+6(-1)=-2+8-6=0.

Since this is 00, the two lines are coplanar.

Now b⃗1×b⃗2=∣i^j^k^123234∣=i^(8−9)−j^(4−6)+k^(3−4)=−i^+2j^−k^\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\2&3&4\end{vmatrix}=\hat i(8-9)-\hat j(4-6)+\hat k(3-4)=-\hat i+2\hat j-\hat k. …

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