# If the straight lines $\dfrac{{x - 1}}{k} = \dfrac{{y - 2}}{2} = \dfrac{{z - 3}}{3}$ and $\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{k} = \dfrac{{z - 1}}{3}$ intersect at a point then the integer K is equal to

A)-2 B)-5 C) 5 D) 2

Last updated date: 23rd Mar 2023

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Answer

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Hint: To solve this we need to know the concept of straight lines and basic factorization of quadratic equations.

Complete step-by-step answer:

Given straight lines are

$\dfrac{{x - 1}}{k} = \dfrac{{y - 2}}{2} = \dfrac{{z - 3}}{3} - - - - - - - - > (1)$

$\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{k} = \dfrac{{z - 1}}{3} - - - - - - - - > (2)$

According to data we know that these two straight lines intersect at a point.

Since we know that if two lines intersect at a point. Then the distance between is equal to zero.

Now by using the above condition let us find the distance between which is equal to zero.

\[\therefore \left[ {\begin{array}{*{20}{c}}

k&2&3 \\

3&k&2 \\

1&1&{ - 2}

\end{array}} \right] = 0\]

$ \Rightarrow k( - 2k - 2) - 2( - 6 - 2) + 3(3 - k) = 0$

$ \Rightarrow 2{k^2} + 5k - 25 = 0$

$ \Rightarrow (2k - 5)(k + 5) = 0$

$ \Rightarrow k = \dfrac{5}{2}, - 5$

Hence the value of integer k is -5.

Note: In this problem it is clearly mentioned that the straight lines intersect at a point .So by using the concept that if two straight lines intersect at a point then the shortest distance between them is zero we had to find the k value which is clearly mentioned as integer.

Complete step-by-step answer:

Given straight lines are

$\dfrac{{x - 1}}{k} = \dfrac{{y - 2}}{2} = \dfrac{{z - 3}}{3} - - - - - - - - > (1)$

$\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{k} = \dfrac{{z - 1}}{3} - - - - - - - - > (2)$

According to data we know that these two straight lines intersect at a point.

Since we know that if two lines intersect at a point. Then the distance between is equal to zero.

Now by using the above condition let us find the distance between which is equal to zero.

\[\therefore \left[ {\begin{array}{*{20}{c}}

k&2&3 \\

3&k&2 \\

1&1&{ - 2}

\end{array}} \right] = 0\]

$ \Rightarrow k( - 2k - 2) - 2( - 6 - 2) + 3(3 - k) = 0$

$ \Rightarrow 2{k^2} + 5k - 25 = 0$

$ \Rightarrow (2k - 5)(k + 5) = 0$

$ \Rightarrow k = \dfrac{5}{2}, - 5$

Hence the value of integer k is -5.

Note: In this problem it is clearly mentioned that the straight lines intersect at a point .So by using the concept that if two straight lines intersect at a point then the shortest distance between them is zero we had to find the k value which is clearly mentioned as integer.

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