If the lines frac{x-1}{2}=frac{y+1}{3}=frac{z | vedclass

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(B) For two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ to inter

Vedclass · Accepted Answer

$\frac{9}{2}$

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(B) For two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ to intersect,the condition is $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight| = 0$. Substituting the given values: $(x_1, y_1, z_1) = (1, -1, 1)$,$(a_1, b_1, c_1) = (2, 3, 4)$,$(x_2, y_2, z_2) = (3, k, 0)$,and $(a_2, b_2, c_2) = (1, 2, 1)$. The determinant becomes: $\left|\begin{array}{ccc} 3-1 & k-(-1) & 0-1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight| = 0$. $\left|\begin{array}{ccc} 2 & k+1 & -1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight| = 0$. Expanding the determinant: $2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$. $2(3-8) - (k+1)(2-4) - 1(4-3) = 0$. $2(-5) - (k+1)(-2) - 1(1) = 0$. $-10 + 2k + 2 - 1 = 0$. $2k - 9 = 0$. $k = \frac{9}{2}$.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is:

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