Question

Find $\lambda $, if the vectors $\overrightarrow a = \,\widehat i + 3\widehat j + \widehat k$, \[\overrightarrow b = \,2\widehat i - \widehat j - \widehat k\] and $\overrightarrow c = \lambda \,\widehat i + 3\widehat k$ are coplanar.

Answer 1

Hint:
If there are three vectors in a 3-D space and their scalar triple product is zero, these three vectors are coplanar.
[a, b, c] = 0, which is basically \[\overrightarrow a .\,(\overrightarrow b \, \times \overrightarrow c )\]. Putting the vectors in the vectors in the formula and solving will give us the value of lambda.

Stepwise solution:
Given:
Three vectors are given
$\overrightarrow a = \,\widehat i + 3\widehat j + \widehat k$
\[\overrightarrow b = \,2\widehat i - \widehat j - \widehat k\]
$\overrightarrow c = \lambda \,\widehat i + 3\widehat k$
Steps:
The vectors are $\overrightarrow a = \,\widehat i + 3\widehat j + \widehat k$, \[\overrightarrow b = \,2\widehat i - \widehat j - \widehat k\], $\overrightarrow c = \lambda \,\widehat i + 3\widehat k$
For the vectors \[\overrightarrow a \,,\,\overrightarrow b \] and $\overrightarrow c $to be coplanar the dot product of a vector with the cross product of the other two vectors must be equal to zero, that is \[\overrightarrow a .\,(\overrightarrow b \, \times \overrightarrow c ) = 0\]. To determine this, we need to find the scalar triple product and solve for lambda ($\lambda $).
If the three vectors are in the form, say
\[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\]
\[\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\]
\[\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k\]
Then we place them as,
$
  \widehat i\,\,\,\,\,\,\,\widehat j\,\,\,\,\,\,\widehat k \\
  {a_1}\,\,\,{a_2}\,\,\,\,{a_3} \\
  {b_1}\,\,\,\,{b_2}\,\,\,\,{b_3} \\
  {c_1}\,\,\,\,{c_2}\,\,\,\,{c_3} \\
$
Now solving the following matrix, we get
$ = \,{a_1}({b_2}\,{c_3}\,\, - \,\,{b_3}{c_2})\, - \,{a_2}({b_1}{c_3}\,\, - \,\,{b_3}{c_1})\,\, + \,\,{a_3}({b_1}{c_2}\, - \,{b_2}{c_1})$
Now putting values of vector a, vector b and vector c in the matrix.
$
  \widehat i\,\,\,\,\,\,\,\widehat j\,\,\,\,\,\,\,\,\widehat k \\
  1\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,1 \\
  2\,\,\,\, - 1\,\,\,\, - 1 \\
  0\,\,\,\,\,\,\,\lambda \,\,\,\,\,\,\,3 \\
$
$ = 1( - 3 + \lambda )\, - \,3(6)\, + \,1(2\lambda )$
Now, we know for the three vectors to be coplanar they should be equal to zero.
Therefore, $1( - 3 + \lambda )\, - \,3(6)\, + \,1(2\lambda ) = 0$
$ \Rightarrow \, - 3 + \lambda - 18 + 2\lambda = 0$
$ \Rightarrow \,3\lambda - 21 = 0$
$\therefore \,\,\,\,\lambda = 7$
Therefore, $\lambda $(lambda) is equal to 7 for the three equations or vectors.

Note:
Three vectors are coplanar if they lie on the same plane. Students must already know that the cross product of two vectors is perpendicular to the plane containing the two vectors. The dot product of two perpendicular vectors on the other hand is zero. So, if the dot product of one of the vectors with the cross product of the other two is zero, then all three vectors must not lie in the same plane. This is true in three-dimensional space.