Question

If A = 2i + k, B = i + j + k and C = 4i – 3j + 7k, then a vector R which satisfies $R \times B = C \times B$ and R.A = 0, is
$\left( A \right) - i - 8j + 2k$
$\left( B \right)i - 8j + 2k$
$\left( C \right)i + 8j + 2k$
$\left( D \right)$ None of these.

Answer 1

Hint: In this particular type of question first assume any vector as R vector, then use the concept that in a dot product the multiplication of same vector group is always 1 and the multiplication of different vector group is zero i.e. $i.i = j.j = k.k = 1$ and \[i.j = j.k = k.i = 0\] so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
A = 2i + k, B = i + j + k and C = 4i – 3j + 7k.
Let, vector R = xi + yj + zk
So R.A is the dot product of R with vector A.
Therefore, R.A = (xi + yj + zk).( 2i + k)
Now as we all know that in dot product the multiplication of same vectors is 1 and multiplication of different vectors is zero, i.e. ($i.i = j.j = k.k = 1$) and (\[i.j = j.k = k.i = 0\]) so use this property we have,
$ \Rightarrow R.A = 2x + 0 + z = 2x + z$................ (1)
Now the cross product of vectors (ai + bj + ck) and (di + ej + fk) is given as
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  a&b&c \\
  d&e&f
\end{array}} \right|$
Now it is given that $R \times B = C \times B$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  x&y&z \\
  1&1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  4&{ - 3}&7 \\
  1&1&1
\end{array}} \right|$
Now expand the determinant we have,
$ \Rightarrow i\left( {y - z} \right) - j\left( {x - z} \right) + k\left( {x - y} \right) = i\left( { - 3 - 7} \right) - j\left( {4 - 7} \right) + k\left( {4 + 3} \right)$
$ \Rightarrow i\left( {y - z} \right) - j\left( {x - z} \right) + k\left( {x - y} \right) = - 10i - j\left( { - 3} \right) + 7k$
So on comparing we have,
$ \Rightarrow y - z = - 10$.................. (2)
$ \Rightarrow x - z = - 3$.................. (3)
$ \Rightarrow x - y = 7$......... (4)
Now add equation (1) and (3) we have,
Therefore, 2x + z + x –z = 0 – 3
$ \Rightarrow 3x = - 3$
$ \Rightarrow x = - 1$
Now from equation (4) we have,
$ \Rightarrow - 1 - y = 7$
$ \Rightarrow - 1 - 7 = y$
$ \Rightarrow y = - 8$
Now from equation (2) we have,
$ \Rightarrow - 8 - z = - 10$
$ \Rightarrow z = 10 - 8 = 2$
So the vector R is
$ \Rightarrow R = - i - 8j + 2k$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of question the key concept we have to remember is that always recall how to take the dot and cross product which is all stated above, then first apply dot product and evaluate the condition as above, then apply the cross product and evaluate the conditions as above then simplify these conditions as above we will get the required vector R.