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Miscellaneous Exercise 6B (Solve) · Q95

Q.Show that lines r⃗=(i^+4j^)+λ(i^+2j^+3k^)\vec{r}=(\hat{i}+4\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}) and r⃗=(3j^−k^)+μ(2i^+3j^+4k^)\vec{r}=(3\hat{j}-\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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a⃗1=i^+4j^\vec a_1=\hat i+4\hat j, b⃗1=i^+2j^+3k^\vec b_1=\hat i+2\hat j+3\hat k; a⃗2=3j^−k^\vec a_2=3\hat j-\hat k, b⃗2=2i^+3j^+4k^\vec b_2=2\hat i+3\hat j+4\hat k.

a⃗2−a⃗1=(0−1,3−4,−1−0)=(−1,−1,−1)\vec a_2-\vec a_1=(0-1,3-4,-1-0)=(-1,-1,-1).

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.

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−1)(−1)+(−1)(2)+(−1)(−1)=1−2+1=0(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=(-1)(-1)+(-1)(2)+(-1)(-1)=1-2+1=0, so the lines are coplanar.

Plane: r⃗⋅(b⃗1×b⃗2)=a⃗1⋅(b⃗1×b⃗2)\vec r\cdot(\vec b_1\times\vec b_2)=\vec a_1\cdot(\vec b_1\times\vec b_2). …

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