Question 31: Motion in 2D & Vectors

Question

Three particles P, Q and R are moving along the vectors $$\overrightarrow A  = \widehat i + \widehat j$$, $$\overrightarrow B  = \widehat j + \widehat k$$ and $$\overrightarrow C  =  - \widehat i + \widehat j$$ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $$\overrightarrow A $$ and $$\overrightarrow B $$. Similarly particle Q is moving normal to the plane which contains vector $$\overrightarrow A $$ and $$\overrightarrow C $$. The angle between the direction of motion of P and Q is $${\cos ^{ - 1}}\left( {{1 \over {\sqrt x }}} \right)$$. Then the value of x is _______________.

Accepted Answer

To solve the problem, we need to determine the vectors normal to the planes containing given vectors and find the angle between those normal vectors.
First, let's find the vectors normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$. The normal vector can be calculated using the cross product:
$$\overrightarrow{A} 	imes \overrightarrow{B} = (\widehat{i} + \widehat{j}) 	imes (\widehat{j} + \widehat{k})$$.
Using the properties of the cross product:
$$\begin{aligned} \overrightarrow{A} 	imes \overrightarrow{B} &= (\widehat{i} 	imes \widehat{j}) + (\widehat{i} 	imes \widehat{k}) + (\widehat{j} 	imes \widehat{j}) + (\widehat{j} 	imes \widehat{k}) \ &= \widehat{k} - \widehat{j} + 0 + \widehat{i} \ &= \widehat{i} - \widehat{j} + \widehat{k}. \end{aligned}$$
So, the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$ is $$\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$$.
Next, let's find the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{C}$$:
$$\overrightarrow{A} 	imes \overrightarrow{C} = (\widehat{i} + \widehat{j}) 	imes (-\widehat{i} + \widehat{j})$$.
Using the properties of the cross product:
$$\begin{aligned} \overrightarrow{A} 	imes \overrightarrow{C} &= (\widehat{i} 	imes -\widehat{i}) + (\widehat{i} 	imes \widehat{j}) + (\widehat{j} 	imes -\widehat{i}) + (\widehat{j} 	imes \widehat{j}) \ &= 0 + \widehat{k} - \widehat{k} + 0 \ &= \widehat{k} - \widehat{k} + 0 \ &= 2 \widehat{k}. \end{aligned}$$
So, the vector normal to the plane containing vectors $$\overrightarrow{A}$$ and $$\overrightarrow{C}$$ is $$\overrightarrow{N}_{2} = 2 \widehat{k}$$. But we just need the direction of this vector, not its magnitude, so we can simplify it to $$\widehat{k}$$.
Now, to find the angle between the two normal vectors $$\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$$ and $$\overrightarrow{N}_{2} = \widehat{k}$$, we use the dot product formula:
$$\cos(	heta) = \frac{\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2}}{||\overrightarrow{N}_{1}|| ||\overrightarrow{N}_{2}||}$$
First, calculate the dot product:
$$\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2} = (\widehat{i} - \widehat{j} + \widehat{k}) \cdot \widehat{k} = 0 + 0 + 1 = 1$$
Next, find the magnitude of the vectors:
$$||\overrightarrow{N}_{1}|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$
$$||\overrightarrow{N}_{2}|| = \sqrt{0^2 + 0^2 + 1^2} = 1$$
Now substitute these into the cosine formula:
$$\cos(	heta) = \frac{1}{\sqrt{3} 	imes 1} = \frac{1}{\sqrt{3}} = \sqrt{\frac{1}{3}}$$
Therefore, $$\cos^{-1}(\sqrt{\frac{1}{3}}) = \cos^{-1}(\frac{1}{\sqrt{3}})$$ indicates $$x = 3$$.
So, the value of $$x$$ is:
3

Motion in 2D & Vectors

Integer / Numeric Answer Type