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Question: In the cartesian plane, \(O\) is the origin of the coordinate axes. A person starts at \(O\) and wal...

In the cartesian plane, OO is the origin of the coordinate axes. A person starts at OO and walks a distance of 33 units in the NORTH-EAST direction and reaches the point P.P. From PP he walks 44 units distance parallel to NORTH-WEST direction and reaches the point Q.Q. Express the vector OQ‾\overline{OQ} in terms of i^\hat{i} and j^.\hat{j}.

Explanation

Solution

North-east is halfway between north and east. So, the angle it makes with the x−x-axis is 45∘.45{}^\circ . North-west is halfway between north and west. So, the angle it makes with the y−y- axis is (90+45)∘=135∘.\left( 90+45 \right){}^\circ =135{}^\circ .

Complete step by step solution:
It is given that the person walks a distance of 33 units in the north-east direction and reaches the point P.P.
Since we are discussing the direction, we need to consider both the x−x-component and y−y-component of the direction. We know that the angle made by the path in which the person walks and the x−x-axis is 45∘.45{}^\circ .
Thus, we will get the vector OP‾=3cos⁡45∘i^+3sin⁡45∘j^.\overline{OP}=3\cos 45{}^\circ \hat{i}+3\sin 45{}^\circ \hat{j}.
And this gives us OP‾=32i^+32j^.\overline{OP}=\dfrac{3}{\sqrt{2}}\hat{i}+\dfrac{3}{\sqrt{2}}\hat{j}.
Also, given that the person walks 44 units from PP in the north-west direction and reaches Q.Q.
Here also we are discussing the direction in which the person walks. And so, we are supposed to consider both the x−x-component and the y−y-component of the direction. The angle made by the path in which the person walks from PP to QQ and the x−x-axis is 135∘.135{}^\circ .
Therefore, we will get PQ‾=4cos⁡135∘i^+4sin⁡135∘j^.\overline{PQ}=4\cos 135{}^\circ \hat{i}+4\sin 135{}^\circ \hat{j}.
And from this, PQ‾=−42i^+42j^.\overline{PQ}=-\dfrac{4}{\sqrt{2}}\hat{i}+\dfrac{4}{\sqrt{2}}\hat{j}.
And now, as per the question, we need to find the vector OQ‾\overline{OQ} in terms of i^\hat{i} and j^,\hat{j}, and so, we add OP‾\overline{OP} and PQ‾\overline{PQ}
That is, OQ‾=OP‾+PQ‾=32i^+32j^−42i^+42j^=−12i^+72j^.\overline{OQ}=\overline{OP}+\overline{PQ}=\dfrac{3}{\sqrt{2}}\hat{i}+\dfrac{3}{\sqrt{2}}\hat{j}-\dfrac{4}{\sqrt{2}}\hat{i}+\dfrac{4}{\sqrt{2}}\hat{j}=-\dfrac{1}{\sqrt{2}}\hat{i}+\dfrac{7}{\sqrt{2}}\hat{j}.
Hence OQ‾=OP‾+PQ‾=12(i^+7j^).\overline{OQ}=\overline{OP}+\overline{PQ}=\dfrac{1}{\sqrt{2}}\left( \hat{i}+7\hat{j} \right).

Note: The half way between the cardinal directions are collectively called the intercardinal directions. And they are north-east, north-west, south-east and south-west. The angles they make with the x−x-axis are NE=45∘,NW=135∘,SW=225∘,SE=315∘.NE=45{}^\circ , NW=135{}^\circ , SW=225{}^\circ , SE=315{}^\circ . Also, remember that sin⁡45∘=cos⁡45∘=12,\sin 45{}^\circ =\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}, cos⁡135∘=cos⁡(180−45)∘=−12,sin⁡135∘=sin⁡(180−45)∘=12.\cos 135{}^\circ =\cos \left( 180-45 \right){}^\circ =-\dfrac{1}{\sqrt{2}}, \sin 135{}^\circ =\sin \left( 180-45 \right){}^\circ =\dfrac{1}{\sqrt{2}}.