Question

If a and b be two perpendicular unit vectors such that $\vec x = \vec b - (\vec a \times \vec x)$, then $|\vec x|$ is equal to
A. 1
B. $\sqrt 2$
C. $\dfrac{1}{{\sqrt 2 }}$
D. $\sqrt 3$

Answer 1

Hint:When the two vectors are perpendicular then their dot product is zero. We will use this property to solve this question. Also, the cross product of a vector with its parallel vector or with itself is zero.

Complete step-by-step answer:
Now, we are given two vectors $\vec a$ and $\vec b$ which are perpendicular to each other. So, the angle between them is $90^\circ$ . So, we get
$\vec a.\vec b = \left| a \right|\left| b \right|\cos 90^\circ$
Now, from trigonometry we know that, $\cos 90^\circ = 0$
Therefore, $\vec a.\vec b = 0$
Now, we have $\vec x = \vec b - (\vec a \times \vec x)$
Taking dot product with $\vec x$ both the sides, we get
$\vec x.\vec x = \vec b.\vec x - (\vec a \times \vec x).\vec x$
Now, $\vec x.\vec x = {\left| {\vec x} \right|^2}$. Also, we can write $(\vec a \times \vec x).\vec x = (\vec x \times \vec x).\vec a$ . As, $\vec x$ is cross multiplied with itself, so the angle between them is $0^\circ$ . Therefore, $(\vec x \times \vec x) = 0$
So, we get
${\left| {\vec x} \right|^2} = \vec b.\vec x$
Taking mod both sides, we get
$||\vec x{|^2}| = |\vec b.\vec x|$
$||\vec x{|^2}| = |\vec b|.|\vec x|$
Now, $\vec b$ is a unit vector. Therefore, $|\vec b| = 1$
So, we get
$|\vec x{|^2} = 1.|\vec x|$
$|\vec x| = 1$
So, option (A) is correct.

Note: Whenever we come up with such types of questions, we will use the properties of dot product and cross product. We will first use the condition given in the question. After that, we will use dot product and cross product according to our need while solving the question. Like in this question, we have used the expression given in the question. After that we have used the property of dot product and cross product. Finally after solving we take mod to find the value of $|\vec x|$ . Dont forget its given in the question that vectors a and b are unit vectors.