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If three vectors 2î - ĵ + k̂, î + 2ĵ - 3k̂ and 3î + λĵ + 5k̂ are co-planar, then λ is:
1. -1
2. -2
3. -3
4. -4

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Correct Answer - Option 4 : -4

Concept:

For three vectors \(\rm \vec A\)\(\rm \vec B\) and \(\rm \vec C\) to be co-planar, the volume of the parallelepiped formed by them must be 0.​ i.e. \(\rm [\vec A\ \vec B\ \vec C]\) = 0.

 

Triple Scalar Product (Box Product): is defined as: \(\rm [\vec A\ \vec B\ \vec C]=\vec A.(\vec B\times\vec C)=\begin{vmatrix} \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3 \\\rm c_1 & \rm c_2 & \rm c_3 \end{vmatrix}\).

 

Calculation:

Let the three vectors be \(\rm \vec A=2\hat i - \hat j + \hat k\)\(\rm \vec B=\hat i +2 \hat j -3 \hat k\) and \(\rm \vec C=3\hat i +\lambda \hat j + 5\hat k\). For the three vectors to be co-planar, their Box Product must be 0.

⇒ \(\rm [\vec A\ \vec B\ \vec C]=0\)

⇒ \(\rm \begin{vmatrix} 2& -1 &\ \ \ 1 \\ \rm 1 &\ \ \ 2 & -3 \\ \rm 3 &\ \ \ \lambda & \ \ \ 5 \end{vmatrix}=0\)

⇒ 2[(2)(5) - (-3)(λ)] + (-1)[(-3)(3) - (1)(5)] + 1[(1)(λ) - (2)(3)] = 0

⇒ 2(10 + 3λ) + (9 + 5) + (λ - 6) = 0

⇒ 20 + 6λ + 8 + λ = 0

⇒ 7λ = -28

⇒ λ = -4.

 

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

  • Dot Product is defined as \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
  • Cross Product is defined as \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\) where \(\rm \vec n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).

 

Volume of a parallelepiped, with vectors \(\rm \vec A\)\(\rm \vec B\) and \(\rm \vec C\) as its sides, is given by the box product of the three vectors.

  • Volume = \(\rm [\vec A\ \vec B\ \vec C]\).

 

For three vectors \(\rm \vec A\)\(\rm \vec B\) and \(\rm \vec C\):

  • Triple Cross Product: is defined as: \(\rm \vec A\times(\vec B\times\vec C)=(\vec A.\vec C)\vec B-(\vec A.\vec B)\vec C\).

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