Question

If the points whose position vectors are
3i-2j -k, 2i+3j-4k, - i +j+2k and
4i +5j +lamda k are coplanar, then show that
146/17

​

Answer 1

SOLUTION

GIVEN

The points whose position vectors are

$3 \hat{ \imath} - 2 \hat{ \jmath} - \hat{k}$

$2 \hat{ \imath} + 3\hat{ \jmath} - 4 \hat{k}$

$- \hat{ \imath} + \hat{ \jmath} + 2 \hat{k}$

$4 \hat{ \imath} + 5 \hat{ \jmath} + \lambda \hat{k}$

are coplanar

TO PROVE

$\displaystyle \: \sf{ \: \lambda = \frac{146}{17} }$

EVALUATION

Let A, B, C, D be the given points and O be the origin

Then

$\vec{OA} = 3 \hat{ \imath} - 2 \hat{ \jmath} - \hat{k}$

$\vec{OB} =2 \hat{ \imath} + 3\hat{ \jmath} - 4 \hat{k}$

$\vec{OC} = - \hat{ \imath} + \hat{ \jmath} + 2 \hat{k}$

$\vec{OD} =4 \hat{ \imath} + 5 \hat{ \jmath} + \lambda \hat{k}$

Now

$\vec{AB} = - \hat{ \imath} + 5 \hat{ \jmath} + 3 \hat{k}$

$\vec{AC} = - 4 \hat{ \imath} + 3 \hat{ \jmath} + 3\hat{k}$

$\vec{AD} = 1 \hat{ \imath} + 7 \hat{ \jmath} + ( \lambda + 1) \hat{k}$

Since the given three points are coplanar

$\displaystyle\begin{vmatrix} -1 & 5 & 3\\ - 4 & 3 & 3 \\ 1 & 7 & ( \lambda + 1) \end{vmatrix} = 0$

$\sf{ \implies \: - (3 \lambda + 3 - 21) - 5( - 4 \lambda - 4 - 3 )- 3( - 28 - 3) = 0}$

$\sf{ \implies \: 17 \lambda - 146 = 0}$

$\sf{ \implies \: 17 \lambda = 146 }$

$\displaystyle \: \sf{ \implies \: \lambda = \frac{146}{17} }$

$\displaystyle \: \sf{ \boxed{ \therefore \: \: \: \: \lambda = \frac{146}{17} } \: \: \: \: }$

Hence proved

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