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CGPET 2021 Solved Paper

Section: Mathematics

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Question : 110 of 150

Marks: +1, -0

If the lines

x-2=y-3=\;\frac{z-4}{-k}

\;\frac{x-1}{k}=\;\frac{y-4}{2}=z-5

are coplanar, then

k

any value
exactly one value
exactly two values
exactly three values

Solution:

L_1=\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}

L_2 = \frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1}

∵

L_1

and

L_2

are coplanar.

∴

(a_2-a_1)\cdot(b_1\times b_2)=0

[(a_2-a_1)b_1b_2]=0

Now,

a_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ and } b_1 = \hat{i} + \hat{j} - \hat{k}

a_2 = \hat{i} + 4\hat{j} + 5\hat{k} \text{ and } b_2 = k\hat{i} + 2\hat{j} + \hat{k}

Now,

[(a_2-a_1)b_1b_2]=0

⇒

\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix}=0

-(1+2k)-1(1+k^2)+1(2-k)=0

⇒

-1-2k-1-k^2+2-k=0

⇒

-k^2-3k=0 \Rightarrow k^2+3k=0

⇒

k(k+3)=0 \Rightarrow k=-3,0

∴ k have exactly two values.

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