Question

If $a = 2i + 2j + k$ , $b = - i + 2j + k$ and $c = 3i + j$ then $a + tb$ this perpendicular to c if t is equal to
A.2
B.4
C.6
D.8

Answer 1

We are given with three vectors and one of which is perpendicular to the sum of the two remaining vectors. So we will use the condition that if two vectors are perpendicular to each other then their dot product is zero. Also remember that the dot product of same unit vectors is 1.

Complete step-by-step answer:
Given that $a + tb$ this perpendicular to c
We know that when two vectors are perpendicular their dot product is zero.
$\Rightarrow (a + tb).c = 0$
Use distributive property,
$\Rightarrow a.c + tb.c = 0$
Rearranging the terms,
$\Rightarrow a.c = - tb.c$
Since we have to find the value of t,
$\Rightarrow - t = \dfrac{{a.c}}{{b.c}}$
Putting the values of a , b and c.
$\Rightarrow - t = \dfrac{{\left( {2i + 2j + k} \right).\left( {3i + j} \right)}}{{\left( { - i + 2j + k} \right).\left( {3i + j} \right)}}$
Multiplying the brackets we get,
$\Rightarrow - t = \dfrac{{2i.3i + 2j.3i + k.3i + 2i.j + 2j.j + k.j}}{{ - i.3i + 2j.3i + k.3i - i.j + 2j.j + k.j}}$
We know that in dot product,

$$i.i = j.j = k.k = 1 \\\ i.j = i.k = j.k = 0 \\\$$

So simplifying the above terms,

$$\Rightarrow - t = \dfrac{{2 \times 3 + 0 + 0 + 0 + 2 + 0}}{{ - 1 \times 3 + 0 + 0 + 0 + 2 + 0}} \\\ \Rightarrow - t = \dfrac{{6 + 2}}{{ - 3 + 2}} \\\ \Rightarrow - t = \dfrac{8}{{ - 1}} \\\ \Rightarrow - t = - 8 \\\$$

Cancelling the minus on the sides,
$\Rightarrow t = 8$
Hence the value of $t = 8$
So the correct option is D.

Note: In this problem students must know two important conditions: Dot product of two vectors is zero if they are perpendicular and dot product of same unit vectors is 1 and of dissimilar vectors is 0. But students may fail at this condition of dissimilar unit vectors product. Remember we are performing dot product here.