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Question: If \[a\] and \[b\] are two non collinear vectors and \[x,y\] are two scalars such that \[\overrighta...

If aa and bb are two non collinear vectors and x,yx,y are two scalars such that ax+by=0\overrightarrow a x + \overrightarrow b y = 0 this implies that:
A. x=y=1x = y = - 1
B. x=y=0x = y = 0
C. x=y=1x = y = 1
D. x=y=ix = y = i

Explanation

Solution

In this question, we will go for option verification and find out for which values of x,yx,y the given two vectors aa and bbare non-collinear. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given aa and bb are two non-collinear vectors and x,yx,y are two scalars.
Also, ax+by=0\overrightarrow a x + \overrightarrow b y = 0
That implies a=yxb..................................(1)\overrightarrow a = - \dfrac{y}{x}\overrightarrow b ..................................\left( 1 \right)
If xx and yy are non-zero, then the two vectors aa and bb are collinear, because a=λb\overrightarrow a = \lambda \overrightarrow b where λ\lambda is scalar as shown in the below figure:

Now, we will go for the option verification to check whether the two vectors aa and bb are collinear or not.
A. By substituting x=y=1x = y = - 1 in equation (1), we have

a=11b a=b  \Rightarrow \overrightarrow a = - \dfrac{{ - 1}}{{ - 1}}\overrightarrow b \\\ \therefore \overrightarrow a = - \overrightarrow b \\\

which is of the form a=λb\overrightarrow a = \lambda \overrightarrow b . So, for the values of x=y=1x = y = - 1, aa and bb are collinear vectors.
B. By substituting x=y=0x = y = 0 in equation (1), we have
a=00b\Rightarrow \overrightarrow a = - \dfrac{0}{0}\overrightarrow b
which is an indeterminate form. So, for the values of x=y=0x = y = 0, aa and bb are non-collinear vectors.
C. By substituting x=y=1x = y = 1 in equation (1), we have

a=11b a=b  \Rightarrow \overrightarrow a = - \dfrac{1}{1}\overrightarrow b \\\ \therefore \overrightarrow a = - \overrightarrow b \\\

which is of the form a=λb\overrightarrow a = \lambda \overrightarrow b . So, for the values of x=y=1x = y = 1, aa and bb are collinear vectors.
D. By substituting x=y=ix = y = i in equation (1), we have

a=iib a=b  \Rightarrow \overrightarrow a = - \dfrac{i}{i}\overrightarrow b \\\ \therefore \overrightarrow a = - \overrightarrow b \\\

which is of the form a=λb\overrightarrow a = \lambda \overrightarrow b . So, for the values of x=y=ix = y = i, aa and bb are collinear vectors.
Therefore, only for the values of x=y=0x = y = 0, the two vectors aa and bb are collinear.
Thus, the correct option is B. x=y=0x = y = 0

So, the correct answer is “Option B”.

Note: We can use any of the given conditions to prove the collinearity for two vectors:
1. Two vectors aa and bb are collinear if there exists a number nn such that a=nb\overrightarrow a = n\overrightarrow b .
2. Two vectors are collinear if relations of their coordinates are equal.
3. Two vectors are collinear if their cross product is equal to the zero vector.