Question

Find a unit vector $ \widehat {c,} $ which is perpendicular to $ \overrightarrow a $ and coplanar with $ \overrightarrow a $ and $ \overrightarrow b $ . $ a = i + j + k,{\text{ b = i + 2j - 6k}}{\text{.}} $
A. $ \dfrac{{2i + 3j - 5k}}{{\sqrt {38} }} $
B. $ \dfrac{{2i - 3j - 5k}}{{\sqrt {38} }} $
C. $ \dfrac{{2i + 3j + 5k}}{{\sqrt {38} }} $
D. $ \dfrac{{ - 2i + 3j - 5k}}{{\sqrt {38} }} $

Answer 1

Hint: The unit vector is also known as the normalized vector. Here we will first suppose the unique vector as the sum of the two given vectors. Then find the values for the unit vector and then place the values in the standard formula for the resultant value.

Complete step-by-step answer:
Let us suppose that the vector is given as $ ai + bj + ck $ .
For the given vector to be coplanar with the vectors $ a = i + j + k{\text{ }} $ and $ {\text{b = i + 2j - 6k}}{\text{.}} $
It can be expressed as –
 $ ai + bj + ck = p(i + j + k) + r(i + 2j - 6k) $
Expand the above equation –
 $ ai + bj + ck = pi + pj + pk + ri + 2rj - 6rk $
Make the pair of like terms –
 $ ai + bj + ck = \underline {pi + ri} + \underline {pj + 2rj} + \underline {pk - 6rk} $
Take common in between the like terms –
 $ ai + bj + ck = (p + r)i + (p + 2r)j + (p - 6r)k $
Compare both the side of the equation –
 $
  a = p + r{\text{ }}....{\text{ (i)}} \\
  b = p + 2r{\text{ }}....{\text{ (ii)}} \\
  c = p - 6r{\text{ }}....{\text{ (iii)}} \;
  $
Now, for the vector $ ai + bj + ck $ to be perpendicular to the vector $ i + j + k $
It can be expressed as –
 $ (ai + bj + ck) \cdot (i + j + k) = 0 $
Simplify the above equation –
 $ \Rightarrow a + b + c = 0{\text{ }}....{\text{ (iv)}} $
Add equations (i), (ii) and (iii) –
We get –
 $ $ $ a + b + c = p + r + p + 2r + p - 6r $
Make pair of the like terms-
 $ a + b + c = \underline {p + p + p} + \underline {r + 2r - 6r} $
Simplify the above equation –
 $ a + b + c = 3p - 3r $
Place the value from the equation (iv)
 $ \Rightarrow 0 = 3p - 3r $
Move term one side to another, also when any term is moved from one side to another sign also changes, positive term changes to negative and vice-versa.
 $ \Rightarrow 3r = 3p $
Common multiple from both the sides of the equation cancels each other. Therefore remove from both the sides.
 $ \Rightarrow p = r $ ..... (v)
Place the values of the above equation in equations (i), (ii) and (iii) –
 $
  a = p + r{\text{ }} \Rightarrow {\text{a = r + r = 2r}} \\
  b = p + 2r\; \Rightarrow {\text{b = r + 2r = 3r}} \\
  c = p - 6r\; \Rightarrow {\text{c = r - 6r = - 5r}} \;
  $
Hence, the required vector is $ (2i + 3j - 5k) $
The magnitude of the above vector is $ = \sqrt {{2^2} + {3^2} + {{( - 5)}^2}} = \sqrt {4 + 9 + 25} = \sqrt {38} $
Therefore, the required unit vector $ \widehat c = \dfrac{{(2i + 3j - 5k)}}{{\sqrt {38} }} $ .
So, the correct answer is “Option A”.

Note: Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative number of positive number always gives positive value. Coplanar vectors are the vectors which lie in the three-dimensional space on the same plane.