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Question

Find the vector equation of the plane passing through the points $$(2, 1, -1)$$ and $$(-1, 3, 4)$$ and perpendicular to the plane $$x - 2y + 4z = 10.$$ Also show that the plane thus obtained contains the line $$\vec{r} = -\hat{i} + 3\hat{j} + 4\hat{k} + \lambda(3 \hat{i} - 2\hat{j} - 5\hat{k}).$$


Solution

Let the equation of plane through $$(2, 1, -1)$$ be
$$a(x - 2) + b(y - 1) + c(z + 1) = 0 ....(i)$$
$$\because (i)$$ passes through $$(-1, 3, 4)$$
$$\therefore a (-1 - 2) + b (3 - 1) + c (4 + 1) = 0$$
$$\Rightarrow -3a + 2b + 5c = 0 ...(ii)$$

Also plane (i) is perpendicular to plane $$x - 2y + 4z = 10$$
$$\Rightarrow \vec{n_1} \perp \vec{n_2} \Rightarrow \vec{n_1}. \vec{n_2} = 0$$
$$\therefore 1a - 2b + 4c = 0 ...(iii)$$

From (ii) and (iii), we get
$$\dfrac{a}{8 + 10} = \dfrac{b}{5 + 12} = \dfrac{c}{6 - 2}  \Rightarrow \dfrac{a}{18} = \dfrac{b}{17} = \dfrac{c}{4} = \lambda (say)$$
$$\Rightarrow a = 18\lambda, b = 17\lambda, c = 4\lambda$$

Putting the value of a, b, c in (i), we get
$$18\lambda (x - 2) + 17\lambda (y - 1) + 4\lambda (z + 1) = 0$$
$$\Rightarrow 18x - 36 + 17y - 17 + 4z + 4 = 0$$
$$\Rightarrow 18x + 17y + 4z = 49$$
$$\therefore$$ Required vector equation of plane is
$$\vec{r}. (18\hat{i} + 17\hat{j} + 4\hat{k}) = 49 ...(iv)$$
Obviously plane (iv) contains the line
$$\vec{r} = (-\hat{i} + 3\hat{j} + 4\hat{k}) + \lambda (3\hat{i} - 2\hat{j} - 5\hat{k}) ....(v)$$
Since, point $$(-\hat{i} + 3\hat{j} + 4\hat{k})$$ satisfy equation (iv) and vector $$(18\hat{i} + 17\hat{j} + 4\hat{k})$$ is perpendicular to, $$(3\hat{i} - 2\hat{j} + 5\hat{k}),$$ as $$(-\hat{i} + 3\hat{j} + 4\hat{k}). (18\hat{i} + 17\hat{j} + 4\hat{k}) = -18 + 51 + 16 = 49$$ and $$(18\hat{i} + 17\hat{j} + 4\hat{k} ). (3\hat{i} - 2\hat{j} - 5\hat{k}) = 54 - 34 - 20 = 0$$
Therefore, (iv) contains line (v).
1640112_1782979_ans_9c37838eff5c49b59cdc299028e62b44.png

Mathematics

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