Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry

Options for the question :-
a. any value
b. exactly one value
c. exactly two values exactly three values​

Answer 1

Answer:

Two lines $\frac{x - x _{1}}{ a_{1} } = \frac{y - y_{1}}{ b_{1} } = \frac{z - z_{1}}{ z_{1}}$

and $\frac{x - x _{2}}{ a_{2} } = \frac{y - y _{2}}{ b_{2} } = \frac{z - z _{2}}{ c_{2} }$

are coplanar if $|\frac{ {x_{2} - x_{1} } }{ \frac{a_{1}}{a_{2}} } \frac{y_{2} - y_{1}}{ \frac{b_{1}}{b_{2}} } \frac{z_{2} - z_{1}}{ \frac{c_{1}}{c_{2}} } | = 0$

where $x_{1} = 2 \: y_{1} = 3 \: z_{1} = 4 \\ x_{2} = 1 \: y_{2} = 4 \: z_{2} = 5 \\ and \\ \: a_{1} = 1 \: b_{1} = 1 \: c_{1} = - k \\ a_{2} = k \: b_{2} = 2 \: c_{2} = 1$

$⟼ |\frac{ {1 - 2 } }{ \frac{1}{k} } \frac{4 - 3}{ \frac{1}{2} } \frac{5- 4} {\frac{ - k}{1}} | = 0$

$⟼ |\frac{ - 1 }{ \frac{1}{k} } \frac{1}{ \frac{1}{2} } \frac{1} {\frac{ - k}{1}} | = 0$

$⟼ \: - 1(1 + 2k) - 1(1 + {k}^{2} ) + 1(2 - k) = 0 \\ ⟼ - 1 - 2k - 1 - {k}^{2} + 2 - k = 0 \\ ⟼ - {k}^{2} - 3k = 0 \\ ⟼k(k + 3) = 0 \\k = 0, -3$

For K = 0 we have

$⟼ |\frac{ { - 1 } }{ \frac{1}{0} } \frac{1 }{ \frac{1}{2} } \frac{1}{ \frac{0}{1} } | = 0 \\ ⟼ - 1(1 - 0) - 1(1 - 0) + 1(2 - 0) = - 1 - 1 + 2 = 0$

For K = - 3 we have

$⟼ |\frac{ { - 1 } }{ \frac{1}{ - 3} } \frac{1 }{ \frac{1}{2} } \frac{1}{ \frac{3}{1} } | = 0 \\ ⟼ - 1(1 - 6) - 1(1 + 9 ) + 1(2 + 3) = 0 \\⟼5 - 10 + 5 = 0 \\ k = - 3$

Hence K = 0 , - 3 are valid .

Answer 2

Answer:k have exactly two values.

Answer in the attachment.