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HCU MCA 2013 Paper

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Question : 71 of 75

Marks: +1, -0

The vectors

\lambda i + j + k,\ i + \lambda j - k \text{ and } 2i - j + \lambda k

are coplanar, if

\text{I.}

\lambda

\text{II.}

\lambda

\frac{1\pm\sqrt{5}}{2}

\text{III.}

\lambda

Which of the following is correct?

$\text{I.}$ and $\text{II.}$
$\text{II.}$ and $\text{III.}$
$\text{I.}$ and $\text{III.}$
Only $\text{I}$

Solution:

(d) Let a =

\lambda i + j + k

b =

i + \lambda j - k

c = 2i - j + λk

If these vectors are coplanar,

then

\;\;\;\;\;\;

[a b c] = 0

⇒

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

Expanding along

R_1,

\lambda (\lambda^{2}-1)-1(\lambda+2)+1(-1-2\lambda)=0

\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\lambda^{3} - \lambda - \lambda - 2 - 1 - 2\lambda = 0

\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda^{3} - 4\lambda - 3 = 0

\Rightarrow\;\;\lambda^{2}(\lambda+1) - \lambda(\lambda+1) - 3(\lambda+1) = 0

\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\lambda+1)(\lambda^{2}-\lambda-3)=0

\therefore\;\;\;\;\;\;\;\;\;\;\;\lambda + 1 = 0

and

\lambda^{2} - \lambda - 3 = 0

\Rightarrow\;\;\;\lambda = -1

and

\lambda = \frac{1\pm\sqrt{1+12}}{2} = \frac{1\pm\sqrt{13}}{2}

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