Question

If $a.\hat i = a.(\hat j + \hat i) = a.(\hat i + \hat j + \hat k)$ , then the value of a is equal to?

Answer 1

Hint: In this question, we have to find the vector $\overrightarrow a$ . To proceed Let vector $\overrightarrow a$ be $x\hat i + y\hat j + z\hat k$ and put it in the given equation to get the answer using the properties of the dot product.

Complete step-by-step answer:
Let $\overrightarrow a = x\hat i + y\hat j + z\hat k$
$\Rightarrow a.\hat i = (x\hat i + y\hat j + z\hat k).\hat i = x$
It’s because we know that in dot product $\hat i.\hat i = 1$ as in dot product we have $\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta$ where $\theta$ is the angle between vectors $\overrightarrow a$ and $\overrightarrow b$ , here value of $\theta$ is ${0^0}$ and we know that $\cos {0^0} = 1$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat j = 0$ as in dot product we have $\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta$ where $\theta$ is the angle between vectors $\overrightarrow a$ and $\overrightarrow b$ , here value of $\theta$ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat k = 0$ as in dot product we have $\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta$ where $\theta$ is the angle between vectors $\overrightarrow a$ and $\overrightarrow b$ , here value of $\theta$ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
$\Rightarrow a.(\hat i + \hat j) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j) = x + y$
And also,
$\Rightarrow a.(\hat i + \hat j + \hat k) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j + \hat k) = x + y + z$
According to the question,
$\Rightarrow a.\hat i = a.(\hat j + \hat i) = a.(\hat i + \hat j + \hat k)$
On substituting the values of each from the above results, we get
$\Rightarrow x = x + y = x + y + z$
On solving the first two we get,
$\Rightarrow x = x + y$
$\Rightarrow y = 0$
Solving the second and third we get,
$\Rightarrow x + y = x + y + z$
$\Rightarrow z = 0$
From above we get x = 1
So, on substituting these in vector a we get
$\Rightarrow a = \hat i$
Hence, we can say that the value of $a = \hat i$

Note- For these types of questions, we have to suppose a vector a in the form of variables at the start to proceed further. Also keep in mind the properties of cross and dot product of vectors for such questions.