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Question: In a game of lawn chess, where pieces are moved between the centers of squares that are each \(1m\) ...

In a game of lawn chess, where pieces are moved between the centers of squares that are each 1m1m on edge, a knight is moved in the following way: (1) two squares forward, one square rightward; (2) two squares leftward, one square forward; (3) two squares forward, one square leftward. What are (a) the magnitude and (b) the angle (relative to forward) of the knight’s overall displacement for the series of three moves?

Explanation

Solution

We solve this question by representing each of the moves taken by the knight in vector form. We represent each of the moves in x,yx,ycoordinates in vector form. Horizontal movements as xx and vertical movements as yy. After representing the moves in their vector form we add the vectors and find its magnitude to get the displacement. To find the overall angle relative to the forward we use the formula of the vertical component of a vector.

Complete step by step solution:
Taking the length of each side of the square as one unit.
Vector representation of each move is
First move is two squares forward and one square rightward
1)d⃗1=2j^+1i^1){\vec d_1} = 2\hat j + 1\hat i
Second move is one square forward and two squares leftward
2)d⃗2=j^−2i^2){\vec d_2} = \hat j - 2\hat i
Third move is two squares forward and one square leftward
3)d⃗3=2j^−1i^3){\vec d_3} = 2\hat j - 1\hat i
Here, the displacement of each move is d⃗1,d⃗2,d⃗3{\vec d_1},{\vec d_2},{\vec d_3}
j^\hat j is the representation of movements in vertical direction and i^\hat i is the representation of movements in horizontal direction.
The sum of these vectors gives us the vector representation of the displacement.
d⃗R=d⃗1+d⃗3+d⃗2{{\vec d}_R} = {{\vec d}_1} + {{\vec d}_3} + {{\vec d}_2}
⇒d⃗R=(2+1+2)j^+(1−2−1)i^=5j^−2i^\Rightarrow {{\vec d}_R} = (2 + 1 + 2)\hat j + (1 - 2 - 1)\hat i = 5\hat j - 2\hat i
To find the magnitude of displacement we find the square root of sum of squares of 5j^−2i^5\hat j - 2\hat i
∣d⃗R∣=22+52=29=5.38m\left| {{{\vec d}_R}} \right| = \sqrt {{2^2} + {5^2}} = \sqrt {29} = 5.38m
Here, magnitude of displacement is ∣d⃗R∣\left| {{{\vec d}_R}} \right|
Hence the magnitude of displacement is 5.38m5.38m
The total displacement of the knight in vertical direction is dv=5j^{d_v} =5\hat j
The magnitude of the vertical component of the displacement is given as ∣d⃗v∣=5| {{{\vec d}_v}} | =5
It can also be represented as,
∣d⃗v∣=∣d⃗R∣cosθ\left| {{{\vec d}_v}} \right| = \left| {{{\vec d}_R}} \right|cos\theta
⇒θ=cos−1∣d⃗v∣∣d⃗R∣\Rightarrow \theta = co{s^{ - 1}}\dfrac{{\left| {{{\vec d}_v}} \right|}}{{\left| {{{\vec d}_R}} \right|}}
⇒θ=cos−1(55.35)\Rightarrow \theta = co{s^{ - 1}}(\dfrac{5}{{5.35}})
⇒θ=21.66∘\Rightarrow \theta = 21.66^\circ
Where θ\theta is the angle between the resultant displacement vector and the vertical component of distance.

Hence the angle (relative to forward) of the knight’s overall displacement for the series of three moves is θ=21.66∘\theta = 21.66^\circ

Note: We can also solve this problem by making a pictorial representation of the moves and using the Pythagoras theorem. And then use the formula of the vector vertical component to find the angle between the displacement and the forward relative to the knight. The units of displacement are meters.