Vectors and Geometry Problems

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### Quadrilateral Types **Problem 1:** What type of a quadrilateral EFGH is it if $\vec{EF} = \vec{HG}$ and $\vec{EH} = \vec{FG}$? **Solution:** If $\vec{EF} = \vec{HG}$, it means that EF is parallel to HG and their lengths are equal. If $\vec{EH} = \vec{FG}$, it means that EH is parallel to FG and their lengths are equal. A quadrilateral with both pairs of opposite sides parallel and equal in length is a **parallelogram**. ### Parallelogram Diagonals **Problem 2:** Find the length of diagonals of a parallelogram having adjacent sides $\vec{a} = 2\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - 3\hat{j}$. **Solution:** Let the adjacent sides be $\vec{a}$ and $\vec{b}$. The diagonals of a parallelogram are given by $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$. Given: $\vec{a} = 2\hat{i} + \hat{j}$ $\vec{b} = \hat{i} - 3\hat{j}$ 1. **First diagonal** $\vec{d_1}$: $\vec{d_1} = (2\hat{i} + \hat{j}) + (\hat{i} - 3\hat{j}) = (2+1)\hat{i} + (1-3)\hat{j} = 3\hat{i} - 2\hat{j}$ Length of $\vec{d_1} = |\vec{d_1}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$ 2. **Second diagonal** $\vec{d_2}$: $\vec{d_2} = (2\hat{i} + \hat{j}) - (\hat{i} - 3\hat{j}) = (2-1)\hat{i} + (1-(-3))\hat{j} = \hat{i} + 4\hat{j}$ Length of $\vec{d_2} = |\vec{d_2}| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$ The lengths of the diagonals are $\sqrt{13}$ and $\sqrt{17}$. ### Right-Angled Triangle **Problem 3:** Show that A(0, 2), B($\sqrt{3}$, -1), C(0, -2) are the vertices of a right-angled triangle. **Solution:** To show it's a right-angled triangle, we need to check if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagorean theorem). Alternatively, we can check if the dot product of any two side vectors is zero. Let's calculate the square of the lengths of the sides: $AB^2 = (\sqrt{3} - 0)^2 + (-1 - 2)^2 = (\sqrt{3})^2 + (-3)^2 = 3 + 9 = 12$ $BC^2 = (0 - \sqrt{3})^2 + (-2 - (-1))^2 = (-\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$ $AC^2 = (0 - 0)^2 + (-2 - 2)^2 = 0^2 + (-4)^2 = 0 + 16 = 16$ Check Pythagorean theorem: $BC^2 + AB^2 = 4 + 12 = 16$ $AC^2 = 16$ Since $BC^2 + AB^2 = AC^2$, the triangle ABC is a right-angled triangle with the right angle at B (opposite to the longest side AC). ### Isosceles Triangle **Problem 4:** Show that A(3, 1), B(-2, -3), C(2, 2) are the vertices of an isosceles triangle. **Solution:** To show it's an isosceles triangle, we need to show that at least two sides have equal lengths. Calculate the square of the lengths of the sides: $AB^2 = (-2 - 3)^2 + (-3 - 1)^2 = (-5)^2 + (-4)^2 = 25 + 16 = 41$ $BC^2 = (2 - (-2))^2 + (2 - (-3))^2 = (2 + 2)^2 + (2 + 3)^2 = 4^2 + 5^2 = 16 + 25 = 41$ $AC^2 = (2 - 3)^2 + (2 - 1)^2 = (-1)^2 + 1^2 = 1 + 1 = 2$ Since $AB^2 = BC^2 = 41$, it means that $AB = BC$. Therefore, the triangle ABC is an isosceles triangle. ### Position Vector Division **Problem 5:** Position vectors of points C and D are $3\hat{i} + 2\hat{j}$ and $\hat{i} - 4\hat{j}$ respectively. Find the position vector of point O dividing the line segment joining C and D in the ratio 2:3 internally. **Solution:** Let the position vectors of C and D be $\vec{c}$ and $\vec{d}$. $\vec{c} = 3\hat{i} + 2\hat{j}$ $\vec{d} = \hat{i} - 4\hat{j}$ The ratio is $m:n = 2:3$. The position vector of point O dividing CD internally in the ratio $m:n$ is given by the section formula: $\vec{o} = \frac{n\vec{c} + m\vec{d}}{m+n}$ $\vec{o} = \frac{3(3\hat{i} + 2\hat{j}) + 2(\hat{i} - 4\hat{j})}{2+3}$ $\vec{o} = \frac{(9\hat{i} + 6\hat{j}) + (2\hat{i} - 8\hat{j})}{5}$ $\vec{o} = \frac{(9+2)\hat{i} + (6-8)\hat{j}}{5}$ $\vec{o} = \frac{11\hat{i} - 2\hat{j}}{5}$ $\vec{o} = \frac{11}{5}\hat{i} - \frac{2}{5}\hat{j}$ The position vector of point O is $\frac{11}{5}\hat{i} - \frac{2}{5}\hat{j}$. ### Midpoint Coordinates **Problem 6:** (2, 4) is mid-point of XY such that X(-1, 3) and Y(x, y). Find the point Y. **Solution:** Let M be the midpoint (2, 4), and the coordinates of X be $(x_1, y_1) = (-1, 3)$, and Y be $(x_2, y_2) = (x, y)$. The midpoint formula is: $M_x = \frac{x_1 + x_2}{2}$ $M_y = \frac{y_1 + y_2}{2}$ For the x-coordinate: $2 = \frac{-1 + x}{2}$ $4 = -1 + x$ $x = 4 + 1 = 5$ For the y-coordinate: $4 = \frac{3 + y}{2}$ $8 = 3 + y$ $y = 8 - 3 = 5$ So, the coordinates of point Y are (5, 5). ### Trisecting a Vector **Problem 7:** If $\vec{OA} = -\hat{i} + 3\hat{j}$ and $\vec{OB} = 6\hat{i} + 2\hat{j}$, find the point trisecting the join of AB. **Solution:** Trisecting a line segment means dividing it into three equal parts. There will be two points that trisect AB. Let these points be P and Q. Let $\vec{a} = \vec{OA} = -\hat{i} + 3\hat{j}$ and $\vec{b} = \vec{OB} = 6\hat{i} + 2\hat{j}$. 1. **Point P (divides AB in ratio 1:2):** $\vec{p} = \frac{2\vec{a} + 1\vec{b}}{1+2} = \frac{2(-\hat{i} + 3\hat{j}) + (6\hat{i} + 2\hat{j})}{3}$ $\vec{p} = \frac{-2\hat{i} + 6\hat{j} + 6\hat{i} + 2\hat{j}}{3}$ $\vec{p} = \frac{4\hat{i} + 8\hat{j}}{3} = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j}$ 2. **Point Q (divides AB in ratio 2:1):** $\vec{q} = \frac{1\vec{a} + 2\vec{b}}{1+2} = \frac{(-\hat{i} + 3\hat{j}) + 2(6\hat{i} + 2\hat{j})}{3}$ $\vec{q} = \frac{-\hat{i} + 3\hat{j} + 12\hat{i} + 4\hat{j}}{3}$ $\vec{q} = \frac{11\hat{i} + 7\hat{j}}{3} = \frac{11}{3}\hat{i} + \frac{7}{3}\hat{j}$ The points trisecting the join of AB are $(\frac{4}{3}, \frac{8}{3})$ and $(\frac{11}{3}, \frac{7}{3})$. ### Collinear Points **Problem 8:** Find x, such that the points P(-1, x), Q(3, 2) and R(7, 3) are collinear. **Solution:** Points are collinear if the slope between any two pairs of points is the same. Slope of PQ = Slope of QR Slope $m = \frac{y_2 - y_1}{x_2 - x_1}$ Slope of PQ: $m_{PQ} = \frac{2 - x}{3 - (-1)} = \frac{2 - x}{4}$ Slope of QR: $m_{QR} = \frac{3 - 2}{7 - 3} = \frac{1}{4}$ For collinearity, $m_{PQ} = m_{QR}$: $\frac{2 - x}{4} = \frac{1}{4}$ $2 - x = 1$ $x = 2 - 1$ $x = 1$ The value of x is 1. ### Parallelogram Vertex **Problem 9:** Find the coordinates of vertex D of a parallelogram ABCD if A(-3, 0), B(1, -2) and C(2, 2) are its three vertices. **Solution:** In a parallelogram ABCD, the diagonals bisect each other. This means the midpoint of AC is the same as the midpoint of BD. Let D be $(x, y)$. Midpoint of AC: $(\frac{-3+2}{2}, \frac{0+2}{2}) = (\frac{-1}{2}, \frac{2}{2}) = (-\frac{1}{2}, 1)$ Midpoint of BD: $(\frac{1+x}{2}, \frac{-2+y}{2})$ Equating the midpoints: $\frac{1+x}{2} = -\frac{1}{2} \implies 1+x = -1 \implies x = -2$ $\frac{-2+y}{2} = 1 \implies -2+y = 2 \implies y = 4$ So, the coordinates of vertex D are (-2, 4). ### Vector Equality **Problem 10:** If $\vec{AB} = \vec{CD}$ and A(0, 2), C(-2, 4), D(-1, 5), then find vertex B. **Solution:** Let B be $(x, y)$. Vector $\vec{AB}$: $\vec{AB} = (x - 0)\hat{i} + (y - 2)\hat{j} = x\hat{i} + (y - 2)\hat{j}$ Vector $\vec{CD}$: $\vec{CD} = (-1 - (-2))\hat{i} + (5 - 4)\hat{j} = (-1 + 2)\hat{i} + 1\hat{j} = 1\hat{i} + 1\hat{j}$ Given $\vec{AB} = \vec{CD}$: $x\hat{i} + (y - 2)\hat{j} = 1\hat{i} + 1\hat{j}$ Equating components: $x = 1$ $y - 2 = 1 \implies y = 3$ So, the coordinates of vertex B are (1, 3). ### Quadrilateral Midpoints **Problem 11:** Vertices of a quadrilateral are U(9, 4), V(-7, 7), W(-4, -7) and X(5, -5). Find midpoints of its sides and prove that the figure obtained by joining mid points consecutively is a parallelogram. **Solution:** Let's find the midpoints of the sides: - **Midpoint of UV (P):** $P_x = \frac{9 + (-7)}{2} = \frac{2}{2} = 1$ $P_y = \frac{4 + 7}{2} = \frac{11}{2} = 5.5$ $P(1, 5.5)$ - **Midpoint of VW (Q):** $Q_x = \frac{-7 + (-4)}{2} = \frac{-11}{2} = -5.5$ $Q_y = \frac{7 + (-7)}{2} = \frac{0}{2} = 0$ $Q(-5.5, 0)$ - **Midpoint of WX (R):** $R_x = \frac{-4 + 5}{2} = \frac{1}{2} = 0.5$ $R_y = \frac{-7 + (-5)}{2} = \frac{-12}{2} = -6$ $R(0.5, -6)$ - **Midpoint of XU (S):** $S_x = \frac{5 + 9}{2} = \frac{14}{2} = 7$ $S_y = \frac{-5 + 4}{2} = \frac{-1}{2} = -0.5$ $S(7, -0.5)$ Now, we need to prove that PQRS is a parallelogram. We can do this by showing that opposite sides are parallel (same slope) or that the diagonals bisect each other. Let's use slopes. Slope of PQ: $m_{PQ} = \frac{0 - 5.5}{-5.5 - 1} = \frac{-5.5}{-6.5} = \frac{11}{13}$ Slope of QR: $m_{QR} = \frac{-6 - 0}{0.5 - (-5.5)} = \frac{-6}{6} = -1$ Slope of RS: $m_{RS} = \frac{-0.5 - (-6)}{7 - 0.5} = \frac{5.5}{6.5} = \frac{11}{13}$ Slope of SP: $m_{SP} = \frac{5.5 - (-0.5)}{1 - 7} = \frac{6}{-6} = -1$ Since $m_{PQ} = m_{RS} = \frac{11}{13}$ (PQ || RS) and $m_{QR} = m_{SP} = -1$ (QR || SP), the quadrilateral PQRS has both pairs of opposite sides parallel. Therefore, the figure obtained by joining the midpoints consecutively is a parallelogram. This is a demonstration of Varignon's Theorem. ### Triangle Midpoints Theorem **Problem 12:** Prove that the line segments joining the midpoints of two sides of a triangle are half the length of the third side. **Solution:** Let a triangle have vertices A, B, C with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively. Let D be the midpoint of AB and E be the midpoint of AC. The position vector of D is $\vec{d} = \frac{\vec{a} + \vec{b}}{2}$. The position vector of E is $\vec{e} = \frac{\vec{a} + \vec{c}}{2}$. The vector $\vec{DE}$ is: $\vec{DE} = \vec{e} - \vec{d} = \frac{\vec{a} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2}$ $\vec{DE} = \frac{1}{2}(\vec{a} + \vec{c} - \vec{a} - \vec{b})$ $\vec{DE} = \frac{1}{2}(\vec{c} - \vec{b})$ The vector $\vec{BC}$ is: $\vec{BC} = \vec{c} - \vec{b}$ Comparing $\vec{DE}$ and $\vec{BC}$, we see that $\vec{DE} = \frac{1}{2}\vec{BC}$. This implies two things: 1. Since $\vec{DE}$ is a scalar multiple of $\vec{BC}$, the line segment DE is parallel to BC ($\vec{DE} || \vec{BC}$). 2. The magnitude of $\vec{DE}$ is half the magnitude of $\vec{BC}$, i.e., $|\vec{DE}| = \frac{1}{2}|\vec{BC}|$. Therefore, the length of the line segment joining the midpoints of two sides of a triangle is half the length of the third side. ### Trapezium Midpoints **Problem 13:** Prove that the joining the midpoints of the two non-parallel sides of a trapezium is parallel to its parallel sides. **Solution:** Let a trapezium ABCD have parallel sides AB and DC, so $\vec{AB} || \vec{DC}$. Let the position vectors of A, B, C, D be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively. Let M be the midpoint of the non-parallel side AD, and N be the midpoint of the non-parallel side BC. Position vector of M: $\vec{m} = \frac{\vec{a} + \vec{d}}{2}$ Position vector of N: $\vec{n} = \frac{\vec{b} + \vec{c}}{2}$ The vector joining the midpoints is $\vec{MN}$: $\vec{MN} = \vec{n} - \vec{m} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{d}}{2}$ $\vec{MN} = \frac{1}{2}(\vec{b} + \vec{c} - \vec{a} - \vec{d})$ $\vec{MN} = \frac{1}{2}((\vec{b} - \vec{a}) + (\vec{c} - \vec{d}))$ $\vec{MN} = \frac{1}{2}(\vec{AB} + \vec{DC})$ Since $\vec{AB}$ and $\vec{DC}$ are parallel, $\vec{AB} = k\vec{DC}$ for some scalar $k$. Then $\vec{MN} = \frac{1}{2}(k\vec{DC} + \vec{DC}) = \frac{1}{2}(k+1)\vec{DC}$. Since $\vec{MN}$ is a scalar multiple of $\vec{DC}$ (and thus also of $\vec{AB}$), $\vec{MN}$ is parallel to $\vec{DC}$ and $\vec{AB}$. Therefore, the line segment joining the midpoints of the two non-parallel sides of a trapezium is parallel to its parallel sides. ### Parallelogram Diagonal Ratio **Problem 14:** Prove that the line segment joining mid points of adjacent sides of a square or a parallelogram divides the corresponding diagonal in the ratio 1:3. **Solution:** Let the parallelogram be ABCD. Let A be the origin (0,0). Let $\vec{AB} = \vec{a}$ and $\vec{AD} = \vec{d}$. Then $\vec{AC} = \vec{a} + \vec{d}$. Let P be the midpoint of AB, and Q be the midpoint of AD. Position vector of P: $\vec{p} = \frac{1}{2}\vec{a}$ Position vector of Q: $\vec{q} = \frac{1}{2}\vec{d}$ The line segment joining the midpoints is PQ. The vector $\vec{PQ} = \vec{q} - \vec{p} = \frac{1}{2}(\vec{d} - \vec{a})$. The diagonal is AC. Consider the triangle APQ and the triangle formed by the diagonal AC. This problem is usually approached by considering the triangle formed by the vertices and the midpoints, and applying Ceva's theorem or Menelaus' theorem, or vector methods. Let the line segment PQ intersect the diagonal AC at point R. We want to show that R divides AC in the ratio 1:3. Any point R on AC can be written as $\vec{r} = (1-t)\vec{A} + t\vec{C} = t(\vec{a} + \vec{d})$ (since A is origin). Any point R on PQ can be written as $\vec{r} = (1-s)\vec{P} + s\vec{Q} = (1-s)\frac{1}{2}\vec{a} + s\frac{1}{2}\vec{d}$. Equating the two expressions for $\vec{r}$: $t(\vec{a} + \vec{d}) = \frac{1-s}{2}\vec{a} + \frac{s}{2}\vec{d}$ Since $\vec{a}$ and $\vec{d}$ are non-collinear (adjacent sides of a parallelogram), we can equate their coefficients: $t = \frac{1-s}{2}$ $t = \frac{s}{2}$ From the second equation, $s = 2t$. Substitute $s$ into the first equation: $t = \frac{1 - 2t}{2}$ $2t = 1 - 2t$ $4t = 1$ $t = \frac{1}{4}$ So, $\vec{r} = \frac{1}{4}(\vec{a} + \vec{d})$. This means R divides AC in the ratio $1:3$ (since $t = \frac{1}{1+3}$). The point R is $\frac{1}{4}$ of the way from A to C, meaning AR : RC = 1 : 3. This applies to both squares and parallelograms.