Question

If the vectors \[4i + 11j + mk,7i + 2j + 6k\] and \[i + 5j + 4k\] are coplanar, then \[{\rm{m}}\] is
A. 38
B. 0
C. 1
D. 10

Answer 1

Hint: If we can express one vector as the linear combination of the other two vectors, then the three vectors supplied are coplanar. Therefore, if we can demonstrate that \[4i + 11j + mk,7i + 2j + 6k\] and \[i + 5j + 4k\] for some value of x and y, we can conclude that the three supplied vectors are coplanar. Any real number may represent x and y's values.

Formula Used:Vectors \[\vec a,\vec b,\vec c\] are coplanar if their scalar triple product is zero.
\[[\vec a\vec b\vec c] = 0\]

Complete step by step solution: \[\vec a = 4\hat i + 11\hat j + m\hat k\]
\[\vec b = 7\hat i + 2\hat j + 6\hat k\]
\[\vec c = \hat i + 5\hat j + 4\hat k\]
We have been already known that vectors \[\vec a,\vec b,\vec c\] are coplanar if their scalar triple product is zero.
That is,
\[[\vec a\vec b\vec c] = 0\]
We know that, the matrix format is,
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Now, we have to write the given vectors in the question as matrix, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}4&{11}&m\\7&2&6\\1&5&4\end{array}} \right| = 0\]
Now, let’s solve the matrix along row, we obtain
\[ \Rightarrow 4(8 - 30) - 11(28 - 6) + m(35 - 2) = 0\]
On subtracting the terms inside the parentheses from the above equation, we get
\[4( - 22) - 11(22) + m(33) = 0\]
On multiplying each term in the above equation, we get
\[ \Rightarrow - 88 - 242 + 33m = 0\]
Now, let’s subtract the numbers\[ - {\rm{88}} - {\rm{242 = }} - {\rm{330}}\], we get
\[{\rm{33m}} - {\rm{330 = 0}}\]
Now, we have to add \[{\rm{330}}\]to either side of the equation, we obtain
\[{\rm{33m}} - {\rm{330 + 330 = 0 + 330}}\]
Now, let’s simplify the above equation, it gives
\[{\rm{33m}} = {\rm{330}}\]
Let’s solve for \[{\rm{m}}\]by moving \[{\rm{33}}\]to the RHS and divide, it gives
\[m = 10\]
Therefore, if the vectors \[4i + 11j + mk,7i + 2j + 6k\] and \[i + 5j + 4k\] are coplanar, then the value of \[{\rm{m}}\] is \[{\rm{10}}\]

Option ‘D’ is correct

Note: It should be emphasized that if any vector is the linear combination of all the other vectors, then not just three vectors but n vectors as well can be coplanar. Additionally, in order to obtain the resultant vector when adding two or more vectors, like terms are added.