Unit 6: Vectors in Plane – Exercise 6.3

Plot \(A(-5,4)\), \(B(-6,1)\) and \(C(-3,1)\) to form a \(\triangle ABC\). Translate \(\triangle ABC\) to \(\triangle A'B'C'\) by translation vector \(8\hat{i} - 6\hat{j}\).

New vertex A':

\[ \begin{aligned} A' &= (-5\hat{i} + 4\hat{j}) + (8\hat{i} - 6\hat{j}) \\ &= 3\hat{i} - 2\hat{j} = (3, -2) \end{aligned} \]

New vertex B':

\[ B' = (-6\hat{i} + \hat{j}) + (8\hat{i} - 6\hat{j}) = 2\hat{i} - 5\hat{j} = (2, -5) \]

New vertex C':

\[ C' = (-3\hat{i} + \hat{j}) + (8\hat{i} - 6\hat{j}) = 5\hat{i} - 5\hat{j} = (5, -5) \]

Hence the coordinates of translated triangle \(\triangle A'B'C'\) are \((3, -2)\), \((2, -5)\), and \((5, -5)\).

Plot \(A(-6, -2)\), \(B(-6, -5)\), \(C(-3, -5)\) and \(D(-3, -2)\) to form a square \(ABCD\). Translate square \(ABCD\) to square \(A'B'C'D'\) by translation vector \(9\hat{i} + 7\hat{j}\).

\[ \begin{aligned} A' &= (-6\hat{i} - 2\hat{j}) + (9\hat{i} + 7\hat{j}) = 3\hat{i} + 5\hat{j} = (3, 5) \\ B' &= (-6\hat{i} - 5\hat{j}) + (9\hat{i} + 7\hat{j}) = 3\hat{i} + 2\hat{j} = (3, 2) \\ C' &= (-3\hat{i} - 5\hat{j}) + (9\hat{i} + 7\hat{j}) = 6\hat{i} + 2\hat{j} = (6, 2) \\ D' &= (-3\hat{i} - 2\hat{j}) + (9\hat{i} + 7\hat{j}) = 6\hat{i} + 5\hat{j} = (6, 5) \end{aligned} \]

Hence the vertices of square \(A'B'C'D'\) are \(A'(3,5)\), \(B'(3,2)\), \(C'(6,2)\), and \(D'(6,5)\).

Plot \(A(-6, 3)\), \(B(-4, 0)\), \(C(-2, 3)\) and \(D(-4, 4)\) to form a kite \(ABCD\). Translate kite \(ABCD\) to kite \(A'B'C'D'\) by translation vector \(6\hat{i} - 6\hat{j}\).

\[ \begin{aligned} A' &= (-6\hat{i} + 3\hat{j}) + (6\hat{i} - 6\hat{j}) = 0\hat{i} - 3\hat{j} = (0, -3) \\ B' &= (-4\hat{i} + 0\hat{j}) + (6\hat{i} - 6\hat{j}) = 2\hat{i} - 6\hat{j} = (2, -6) \\ C' &= (-2\hat{i} + 3\hat{j}) + (6\hat{i} - 6\hat{j}) = 4\hat{i} - 3\hat{j} = (4, -3) \\ D' &= (-4\hat{i} + 4\hat{j}) + (6\hat{i} - 6\hat{j}) = 2\hat{i} - 2\hat{j} = (2, -2) \end{aligned} \]

Hence the vertices of kite \(A'B'C'D'\) are \(A'(0,-3)\), \(B'(2,-6)\), \(C'(4,-3)\), and \(D'(2,-2)\).

The coordinates of \(A\), \(B\) and \(D\) are \((1, 2)\), \((6, 3)\) and \((2, 8)\) respectively. Find the coordinates of \(C\) by using vector method if \(ABCD\) is a parallelogram.

Let coordinates of \(C = C(x,y)\). Since \(ABCD\) is a parallelogram, \(\overline{AB} = \overline{DC}\).

\[ \begin{aligned} \overline{AB} &= \overline{DC} \\ (6\hat{i} + 3\hat{j}) - (\hat{i} + 2\hat{j}) &= (x\hat{i} + y\hat{j}) - (2\hat{i} + 8\hat{j}) \\ 5\hat{i} + \hat{j} &= (x-2)\hat{i} + (y-8)\hat{j} \end{aligned} \]

Equating components:

\[ x - 2 = 5 \implies x = 7,\quad y - 8 = 1 \implies y = 9 \]

Hence coordinates of \(C = C(7, 9)\).

In parallelogram \(ABCD\), \(\overline{AB} = 6\hat{i} + 2\hat{j}\) and \(\overline{DC} = -6\hat{i} - 2\hat{j}\). Show that the opposite sides are equal in magnitude and parallel.

Points \(A(1,2)\), \(B(4,6)\) and \(C(7,2)\) form a triangle. Check whether triangle \(ABC\) is an isosceles by using vector magnitude.

\[ \begin{aligned} \overline{AB} &= 3\hat{i} + 4\hat{j},\quad |\overline{AB}| = 5 \\ \overline{BC} &= 3\hat{i} - 4\hat{j},\quad |\overline{BC}| = 5 \\ \overline{AC} &= 6\hat{i} + 0\hat{j},\quad |\overline{AC}| = 6 \end{aligned} \]

Since \(|\overline{AB}| = |\overline{BC}| = 5\), triangle \(ABC\) is isosceles.

Use vectors to show that \(PQRS\) is a parallelogram, where \(P(1,2)\), \(Q(5,2)\), \(R(7,6)\) and \(S(3,6)\).

\[ \begin{aligned} \overline{PQ} &= 4\hat{i} + 0\hat{j},\quad \overline{SR} = 4\hat{i} + 0\hat{j} \\ \overline{PS} &= 2\hat{i} + 4\hat{j},\quad \overline{QR} = 2\hat{i} + 4\hat{j} \end{aligned} \]

Since \(\overline{PQ} = \overline{SR}\) and \(\overline{PS} = \overline{QR}\), \(PQRS\) is a parallelogram.

Use vectors to show that triangle \(XYZ\) is an isosceles, where \(X(0,0)\), \(Y(2,0)\) and \(Z(1,3)\).

\[ \begin{aligned} \overline{XY} &= 2\hat{i} + 0\hat{j},\quad |\overline{XY}| = 2 \\ \overline{YZ} &= -\hat{i} + 3\hat{j},\quad |\overline{YZ}| = \sqrt{10} \\ \overline{XZ} &= \hat{i} + 3\hat{j},\quad |\overline{XZ}| = \sqrt{10} \end{aligned} \]

Since \(|\overline{YZ}| = |\overline{XZ}| = \sqrt{10}\), triangle \(XYZ\) is isosceles.

A ball is projected with initial velocity vector \(\bar{v}_0 = 10\hat{i} + 20\hat{j}\). The horizontal component is in the x-direction and gravity is \(\bar{g} = 0\hat{i} - 10\hat{j}\). Find the maximum height and horizontal range.

Given: \(v_x = 10\), \(v_{iy} = 20\), \(g = -10\)

Maximum Height (h): At maximum height, \(v_{fy} = 0\).

\[ \begin{aligned} 2gh &= v_f^2 - v_i^2 \\ 2(-10)h &= 0^2 - 20^2 \\ -20h &= -400 \\ h &= 20 \end{aligned} \]

Horizontal Range (R): Time to reach max height: \(v_f = v_i + gt\)

\[ 0 = 20 - 10t \implies t = 2 \]

Total time of flight: \(T = 2t = 4\)

\[ R = v_x \times T = 10 \times 4 = 40 \]

Hence, Maximum Height = 20 units, Horizontal Range = 40 units.

A car enters a loop with velocity vector \(\bar{v} = 30\hat{j}\) and exits with velocity \(\bar{v}' = 30\hat{i}\). What is the change in velocity vector?

An aeroplane has airspeed \(\bar{v}_p = 20\hat{i}\) and there is a crosswind \(\bar{v}_w = 50\hat{j}\). Find the resultant velocity and its magnitude.