The vector equation of the plane passing thro | vedclass

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Question

(A) The equation of a plane passing through the line of intersection of the planes $P_1: r \cdot n_1 = d_1$ and $P_2: r \cdot n_2 = d_2$ is given by $

Vedclass · Accepted Answer

$r \cdot (i + 9j + 11k) = 0$

Vedclass · Answer

(A) The equation of a plane passing through the line of intersection of the planes $P_1: r \cdot n_1 = d_1$ and $P_2: r \cdot n_2 = d_2$ is given by $(r \cdot n_1 - d_1) + \lambda (r \cdot n_2 - d_2) = 0$.Here,$n_1 = i + 3j - k$,$d_1 = 0$,$n_2 = j + 2k$,and $d_2 = 0$.So,the equation is $(r \cdot (i + 3j - k)) + \lambda (r \cdot (j + 2k)) = 0$ ..... $(i)$This plane passes through the point $(2, 1, -1)$,which corresponds to the position vector $a = 2i + j - k$.Substituting $r = 2i + j - k$ into equation $(i)$:$((2i + j - k) \cdot (i + 3j - k)) + \lambda ((2i + j - k) \cdot (j + 2k)) = 0$$(2(1) + 1(3) + (-1)(-1)) + \lambda (2(0) + 1(1) + (-1)(2)) = 0$$(2 + 3 + 1) + \lambda (0 + 1 - 2) = 0$$6 + \lambda (-1) = 0 \Rightarrow \lambda = 6$Substituting $\lambda = 6$ back into equation $(i)$:$r \cdot (i + 3j - k) + 6(r \cdot (j + 2k)) = 0$$r \cdot (i + 3j - k + 6j + 12k) = 0$$r \cdot (i + 9j + 11k) = 0$Thus,the correct option is $A$.

The vector equation of the plane passing through the point $(2, 1, -1)$ and the line of intersection of the planes $r \cdot (i + 3j - k) = 0$ and $r \cdot (j + 2k) = 0$ is:

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