I want to solve 4th---Page: 484
Title: Exercise 19.1
Mathematical Expression/Proof Snippet above Exercise:
$\vec{AD} = \frac{1}{2}\vec{AB} = \frac{1}{2}\vec{b}$,
$\vec{AE} = \frac{1}{2}\vec{AC} = \frac{1}{2}\vec{c}$
$\therefore \vec{DE} = \vec{AE} - \vec{AD} = \frac{1}{2}\vec{c} - \frac{1}{2}\vec{b} = \frac{1}{2}(\vec{c}-\vec{b}) = \frac{1}{2}\vec{BC}$
Hence $\vec{DE}$ is half of $\vec{BC}$ and parallel to it.
Problem 1:
Let $ABCD$ be a quadrilateral, prove that
$\vec{BA} + \vec{BC} + \vec{CD} + \vec{DA} = 2\vec{BA}$
(UP 1999)
Problem 2:
From adjoining figure find the value of $\vec{AD}$.
Chart Description for Problem 2 (Fig. 19.17):
Type: Geometric figure (Quadrilateral)
Elements:
- Points: A, B, C, D forming a quadrilateral.
- Lines: Segments AB, BC, CD, DA.
- Vectors: $\vec{AB}$ labeled as $\vec{a}$, $\vec{BC}$ labeled as $\vec{b}$, $\vec{DC}$ labeled as $\vec{c}$.
- Relative Position: Quadrilateral ABCD. Vectors $\vec{a}$ along AB, $\vec{b}$ along BC, $\vec{c}$ along DC. Point A is lower left, B is lower right, C is upper right, D is upper left.
Figure Label: Fig. 19.17
Problem 3:
Find the sum of $\vec{AB}$ and $\vec{BC}$ in $\triangle ABC$ of adjoining figure.
Chart Description for Problem 3 (Fig. 19.18):
Type: Geometric figure (Triangle)
Elements:
- Points: A, B, C forming a triangle.
- Lines: Segments AB, BC, CA.
- Relative Position: Triangle ABC. Point A is left, B is upper, C is right.
Figure Label: Fig. 19.18
Problem 4:
If $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}$ then find the value of $\vec{BA}$.
Problem 5:
Find the value of $\vec{OA} + \vec{OB}$ in the adjoining figure.
Chart Description for Problem 5 (Fig. 19.19):
Type: Geometric figure (Square/Rectangle - appears square-like but sides are labeled with vectors)
Elements:
- Points: O, A, B, C forming a shape that appears to be a rectangle with origin O.
- Lines: Segments OA, AB, BC, CO.
- Vectors: $\vec{OA}$ labeled as $\vec{a}$, $\vec{AB}$ labeled as $\vec{d}$, $\vec{BC}$ labeled as $\vec{b}$, $\vec{OC}$ labeled as $\vec{c}$.
- Relative Position: Point O is lower left, A is lower right, B is upper right, C is upper left. $\vec{a}$ is along OA, $\vec{d}$ is along AB, $\vec{b}$ is along BC, $\vec{c}$ is along OC.
Figure Label: Fig. 19.19
Problem 6:
The position vectors of four points $A, B, C, D$ are $\vec{a}, \vec{b}, 2\vec{a} + 3\vec{b}, \vec{a} - 2\vec{b}$, represent the vectors $\vec{AC}, \vec{DB}, \vec{BC}$ and $\vec{CA}$ in terms of $\vec{a}$ and $\vec{b}$.
Problem 7:
In the adjoining figure, $\vec{OA} = \vec{a}, \vec{AB} = \vec{b}$; then find the value of $\vec{OB}$ and $\vec{CD}$ where $C$ and $D$ are the mid points of $OA$ and $AB$ respectively.
Problem 8:
$ABCD$ is a quadrilateral $\vec{AC}$ and $\vec{BD}$ are its diagonals. Prove that $\vec{AB} + \vec{DC} = \vec{AC} + \vec{DB}$
Problem 9:
If $\vec{a}$ and $\vec{b}$ are the vectors determined by two adjacent sides of a regular hexagon, find the vectors determining the other sides taken in order.
Problem 10:
Let $G$ be the centre of the triangle $ABC$, prove that $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.
Problem 11:
The diagonals of a parallelogram $ABCD$ intersect at $E$. If $O$ be any point, prove that $\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OE}$.
Problem 12:
If $\vec{a}$ and $\vec{b}$ are the position vectors of the points $A$ and $B$ respectively, find the position vector of the point $C$ on the extension of $AB$ such that $\vec{AC} = 3\vec{AB}$ and the point $D$ on the extension of $BA$ such that $\vec{BD} = 2\vec{BA}$.
Problem 13:
$D, E, F$ are the mid points of the sides $BC, CA, AB$ of a $\triangle ABC$. Prove that for any vector $O$,
$\vec{OA} + \vec{OB} + \vec{OC} = \vec{OD} + \vec{OE} + \vec{OF}$.
Problem 14:
Let $ABCDE$ be a pentagon. Forces represented by $\vec{AB}, \vec{BC}, \vec{CD}, \vec{DE}$ and $\vec{AC}$ act on a particle. Prove that the resultant force is $3\vec{AC}$.
Problem 15:
Prove that in a hexagon, the sum of the vectors from the origin is zero.
Chart Description (Fig. 19.20):
Type: Geometric figure (Quadrilateral with diagonals or two adjacent triangles)
Elements:
- Points: O, C, A, D. Appears to be two triangles OAD and OCD joined at OD, or possibly a kite shape.
- Lines: Segments OC, CD, DA, AO, OD.
- Relative Position: Point O is lower left, C is upper left, A is lower right, D is upper right. Line OD is a diagonal. Line CA appears to be another diagonal.
Figure Label: Fig. 19.20
Note: This figure does not seem to be directly referenced by problem numbers 1-15 in the visible text. It might be related to an example not shown.
Hint to Selected Problems:
Problem 7 Hint:
$\vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + \vec{b}$
If C and D are the mid points of OA and AB, then
$\vec{OC} = \frac{\vec{OA}}{2} = \frac{\vec{a}}{2}$, $\vec{AD} = \frac{\vec{AB}}{2} = \frac{\vec{b}}{2}$
$\therefore \vec{CD} = \vec{CA} + \vec{AD} = -\vec{AC} + \vec{AD}$.
There seems to be a typo in the hint calculation for $\vec{CD}$. The calculation shown is $\vec{CD} = \vec{CA} + \vec{AD} = \frac{\vec{a}}{2} + \frac{\vec{b}}{2} = \frac{1}{2}(\vec{a}+\vec{b})$. However, from the figure description of Fig 19.19 (which seems relevant to problem 7 based on vector labels $\vec{a}$ and $\vec{b}$ for OA and AB), C is on OC (labeled $\vec{c}$) and D is on AB (labeled $\vec{d}$). The hint relates to C being the midpoint of OA and D being the midpoint of AB. Let's assume the hint's setup is correct for problem 7, regardless of the figure labels. The hint calculates $\vec{OC} = \vec{a}/2$ and $\vec{AD} = \vec{b}/2$. Then calculates $\vec{CD}$ as $\vec{CA} + \vec{AD}$. $\vec{CA} = \vec{OA} - \vec{OC} = \vec{a} - \vec{a}/2 = \vec{a}/2$. So $\vec{CD} = \vec{a}/2 + \vec{b}/2 = \frac{1}{2}(\vec{a}+\vec{b})$. This matches the calculation in the image.
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Let's solve Problem 4. We are given that vector OA equals vector a, and vector OB equals vector b. We need to find vector BA. Using the vector subtraction formula, vector BA equals vector OA minus vector OB. Substituting the given values, we get vector BA equals vector a minus vector b. This is our final answer.
Now let's solve Problem 2. We have a quadrilateral ABCD where vector AB equals vector a, vector BC equals vector b, and vector DC equals vector c. We need to find vector AD. To go from A to D, we can follow the path A to B to C to D. So vector AD equals vector AB plus vector BC plus vector CD. Since vector CD is the opposite of vector DC, we have vector CD equals negative vector c. Therefore, vector AD equals vector a plus vector b minus vector c.
Let's solve Problem 3. In triangle ABC, we need to find the sum of vector AB and vector BC. Using vector addition, when we add vector AB and vector BC, we apply the triangle rule. Vector AB takes us from point A to point B, then vector BC takes us from point B to point C. The result is the same as going directly from A to C, which is vector AC. Therefore, vector AB plus vector BC equals vector AC.
Now let's prove Problem 1. We need to show that in quadrilateral ABCD, the sum of vectors BA, BC, CD, and DA equals 2 times vector BA. We start by grouping the vectors strategically. We can group BC and CD together, which gives us BD by the triangle rule. So our expression becomes BA plus BD plus DA. Then we group BD and DA, which gives us BA again by the triangle rule. Therefore, we have BA plus BA, which equals 2 times BA. This completes our proof.
To summarize what we've learned: Vector subtraction uses the formula BA equals OA minus OB when position vectors are given. The path method allows us to find vectors by following a route through connected points. The triangle rule shows that adding consecutive vectors gives the direct path. These vector algebra techniques provide systematic approaches to solving geometric problems.