Vector

The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by

$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$

i.e. $\theta = \cos^{-1}\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)$, where $\vec{a}\cdot\vec{b}$ is the dot product of the two vectors.

Let $\vec{a} = 3\vec{i} + 4\vec{j}$ and $\vec{b} = 8\vec{i} - 6\vec{j}$.

Dot product: $\vec{a}\cdot\vec{b} = (3)(8) + (4)(-6) = 24 - 24 = 0$

Since $\vec{a}\cdot\vec{b} = 0$, the vectors are perpendicular.

$\therefore \text{Angle between them} = 90^\circ$

Given: A parallelogram $EFGH$ with $\vec{EF} = \vec b$ and $\vec{EH} = \vec d$.

To prove: The diagonals $EG$ and $FH$ bisect each other.

Proof:

Take $E$ as the origin. Then the position vectors of the vertices are

$\vec e = \vec 0,\qquad \vec f = \vec b,\qquad \vec h = \vec d$

Since $EFGH$ is a parallelogram, $\vec{FG} = \vec{EH} = \vec d$, so

$\vec g = \vec f + \vec{FG} = \vec b + \vec d$

Midpoint of diagonal $EG$:

$\frac{\vec e + \vec g}{2} = \frac{\vec 0 + (\vec b + \vec d)}{2} = \frac{\vec b + \vec d}{2}$

Midpoint of diagonal $FH$:

$\frac{\vec f + \vec h}{2} = \frac{\vec b + \vec d}{2}$

The two midpoints have the same position vector, so they are the same point. Hence the diagonals $EG$ and $FH$ intersect at their common midpoint, i.e. they bisect each other. $\blacksquare$

Given: Right-angled triangle $GHI$ with $\angle H=90^\circ$; $M$ is the midpoint of the hypotenuse $GI$.

To prove: $MG=MH=MI$.

Proof: Take $H$ as origin and let $\vec{HG}=\vec g,\ \vec{HI}=\vec i$. Since $\angle H=90^\circ$, $\vec g\cdot\vec i=0$.

$M$ is the midpoint of $GI$, so $\vec{HM}=\dfrac{\vec g+\vec i}{2}$.

$MH^2=\frac14|\vec g+\vec i|^2=\frac14\left(|\vec g|^2+2\vec g\cdot\vec i+|\vec i|^2\right)=\frac14\left(|\vec g|^2+|\vec i|^2\right)$

$MG^2=\left|\vec{HM}-\vec g\right|^2=\frac14|\vec i-\vec g|^2=\frac14\left(|\vec i|^2-2\vec g\cdot\vec i+|\vec g|^2\right)=\frac14\left(|\vec g|^2+|\vec i|^2\right)$

$MI^2=\left|\vec{HM}-\vec i\right|^2=\frac14|\vec g-\vec i|^2=\frac14\left(|\vec g|^2+|\vec i|^2\right)$

Hence $MH^2=MG^2=MI^2$, so $MH=MG=MI$: the midpoint of the hypotenuse is equidistant from the three vertices. $\blacksquare$

To prove: The diagonals $MO$ and $NP$ of rhombus $MNOP$ are perpendicular.

Proof: Take $M$ as origin and let $\vec{MN}=\vec a,\ \vec{MP}=\vec b$. Since all sides of a rhombus are equal, $|\vec a|=|\vec b|$.

Since $\vec{NO}=\vec{MP}=\vec b$,

$\vec{MO}=\vec{MN}+\vec{NO}=\vec a+\vec b,\qquad \vec{NP}=\vec{MP}-\vec{MN}=\vec b-\vec a.$

Then

$\vec{MO}\cdot\vec{NP}=(\vec a+\vec b)\cdot(\vec b-\vec a)=\vec a\cdot\vec b-\vec a\cdot\vec a+\vec b\cdot\vec b-\vec b\cdot\vec a=|\vec b|^2-|\vec a|^2=0.$

The dot product is zero, so $\vec{MO}\perp\vec{NP}$. Hence the diagonals of a rhombus are perpendicular to each other. $\blacksquare$

$\vec a\cdot\vec b=x_1x_2+y_1y_2$

$(\vec{j})^2 = \vec{j}\cdot\vec{j} = |\vec{j}|^2 = 1^2 = 1$

So, the value is 1.

Given: $\vec{a}\cdot\vec{b}=48$, $|\vec{a}|=6\sqrt2$, $|\vec{b}|=8$

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.

Using the dot product formula, $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$

$48=(6\sqrt2)(8)\cos\theta$

$48=48\sqrt2\cos\theta$

$\cos\theta=\frac{48}{48\sqrt2}=\frac1{\sqrt2}$

$\therefore \theta=45^\circ$

So, the angle between $\vec{a}$ and $\vec{b}$ is $\boxed{45^\circ}$.

If $\vec{a}$ and $\vec{b}$ are perpendicular, the angle between them is $90^\circ$. So

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos 90^\circ = 0$

Value of $\vec{a} \cdot \vec{b} = 0$

Given: $|\vec{a}| = 4$, $|\vec{b}| = 6$, $\vec{a}\cdot\vec{b} = 12$

Using formula, $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$

$= \frac{12}{4 \times 6} = \frac{12}{24} = \frac{1}{2}$

$\therefore \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$

So, the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$.

Given: $PQ$ is a diameter of a semicircle with centre $O$, and $M$ is a point on the arc.

To prove: $\angle PMQ=90^\circ$.

Proof: Take $O$ as origin and let $\vec{OP}=\vec p,\ \vec{OQ}=\vec q,\ \vec{OM}=\vec m$.

$O$ is the midpoint of $PQ$, so $\vec q=-\vec p$. Also $M$ is on the arc, so $|\vec m|=|\vec p|$ (the radius).

$\vec{MP}=\vec p-\vec m,\qquad \vec{MQ}=\vec q-\vec m=-\vec p-\vec m$

$\vec{MP}\cdot\vec{MQ}=(\vec p-\vec m)\cdot(-\vec p-\vec m)=-|\vec p|^2-\vec p\cdot\vec m+\vec m\cdot\vec p+|\vec m|^2=|\vec m|^2-|\vec p|^2=0.$

So $\vec{MP}\perp\vec{MQ}$, i.e. $\angle PMQ=90^\circ$. $\blacksquare$

If $\vec{a}\cdot\vec{b} = 0$, then $\vec{a}$ and $\vec{b}$ are perpendicular to each other, i.e. $\vec{a} \perp \vec{b}$ (provided both are non-zero vectors).

Given: $\vec{a}\cdot\vec{b}=\frac{15}{2},\quad |\vec{a}|=5,\quad |\vec{b}|=3$

Using the dot product formula, $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$

$\frac{15}{2}=5\times 3\times \cos\theta$

$\frac{15}{2}=15\cos\theta$

$\cos\theta=\frac{1}{2}$

$\theta=60^\circ$

So, the angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$.

Let $ABCD$ be a quadrilateral. Let $P,Q,R,S$ be the midpoints of the adjacent sides $AB,BC,CD,DA$ respectively.

Take the position vectors of $A,B,C,D$ as $\vec a,\vec b,\vec c,\vec d$.

Since $P$ is the midpoint of $AB$, $\vec p=\frac{\vec a+\vec b}{2}$ Similarly, $\vec q=\frac{\vec b+\vec c}{2},\qquad \vec r=\frac{\vec c+\vec d}{2},\qquad \vec s=\frac{\vec d+\vec a}{2}$

Now, $\vec{PQ}=\vec q-\vec p=\frac{\vec b+\vec c}{2}-\frac{\vec a+\vec b}{2}=\frac{\vec c-\vec a}{2}$

Also, $\vec{SR}=\vec r-\vec s=\frac{\vec c+\vec d}{2}-\frac{\vec d+\vec a}{2}=\frac{\vec c-\vec a}{2}$

Hence, $\vec{PQ}=\vec{SR}$

So the opposite sides $PQ$ and $SR$ are equal and parallel.

Therefore, the quadrilateral $PQRS$ formed by joining the midpoints of the adjacent sides of quadrilateral $ABCD$ is a parallelogram. $\blacksquare$

$\vec{a}$ and $\vec{b}$ are parallel and in the same direction.

Since $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$, the given condition means $\cos\theta=1$, so $\theta=0^\circ$.

Since $\vec{a}\perp\vec{b}$, $\vec{a}\cdot\vec{b}=0$:

$4(-m)+3(-8)=0$ $-4m-24=0\Rightarrow m=-6$

$\boxed{m=-6}$

$\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$

Perpendicular vectors have zero scalar product.

$\vec a\cdot\vec b=(5)(3)+(3)(-m)=15-3m=0$ $m=5$

Since $P$ is the mid-point of $AB$,

$\overrightarrow{OP}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}\ \Rightarrow\ \overrightarrow{OB}=2\overrightarrow{OP}-\overrightarrow{OA}$

$\overrightarrow{OB}=2(5\vec i+2\vec j)-(4\vec i+5\vec j)=6\vec i-\vec j$

So the position vector of $B$ is $6\vec i-\vec j$.

$\vec a\cdot\vec b=x_1x_2+y_1y_2$

Since $\vec a+\vec b+\vec c=\vec0$, we have $\vec c=-(\vec a+\vec b)$. Let $\theta$ be the angle between $\vec a$ and $\vec b$.

$|\vec c|^2=|\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cos\theta$

$(3\sqrt2)^2=6^2+(3\sqrt2)^2+2(6)(3\sqrt2)\cos\theta$ $18=36+18+36\sqrt2\cos\theta$ $\cos\theta=-\frac{36}{36\sqrt2}=-\frac1{\sqrt2}$

Hence $\theta=135^\circ$.

Let $BE$ and $CF$ be the medians of $\triangle ABC$, where $E$ is the midpoint of $AC$ and $F$ is the midpoint of $AB$. Take $A$ as the origin, with $\overrightarrow{AB}=\vec b$ and $\overrightarrow{AC}=\vec c$.

Then $\overrightarrow{AE}=\tfrac12\vec c$ and $\overrightarrow{AF}=\tfrac12\vec b$, so

Squaring the magnitudes:

Given $BE=CF$, so $|\overrightarrow{BE}|^2=|\overrightarrow{CF}|^2$:

Hence $AB=AC$, so $\triangle ABC$ is isosceles. $\blacksquare$

Since $M$ is the midpoint of $AB$, $\overrightarrow{AM}=\tfrac12\overrightarrow{AB}$.

Now $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\vec b-\vec a$.

By the triangle law in $\triangle OAM$,

$\vec m=\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\vec a+\tfrac12(\vec b-\vec a)=\tfrac12\vec a+\tfrac12\vec b=\tfrac12(\vec a+\vec b).\ \blacksquare$

Let $OABC$ be a rectangle with $O$ as origin, $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OC}=\vec c$. Since the adjacent sides are perpendicular, $\vec a\cdot\vec c=0$. Also $\overrightarrow{CB}=\vec a$, so $\overrightarrow{OB}=\vec a+\vec c$.

The diagonals are $OB$ and $AC$:

$\overrightarrow{OB}=\vec a+\vec c,\qquad \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\vec c-\vec a.$

$|\overrightarrow{OB}|^2=|\vec a|^2+2\,\vec a\cdot\vec c+|\vec c|^2=|\vec a|^2+|\vec c|^2$ $|\overrightarrow{AC}|^2=|\vec c|^2-2\,\vec a\cdot\vec c+|\vec a|^2=|\vec a|^2+|\vec c|^2$

So $|\overrightarrow{OB}|=|\overrightarrow{AC}|$: the diagonals of a rectangle are equal. $\blacksquare$

(c) $0$

The two sides forming the right angle, $\overrightarrow{BA}$ and $\overrightarrow{BC}$, are perpendicular to each other. For perpendicular vectors $\theta = 90°$, so the scalar product is $|\overrightarrow{BA}||\overrightarrow{BC}|\cos 90° = 0$.

(a) Let the position vectors of $A$ and $B$ be $\vec a$ and $\vec b$.
Since $D$ is the midpoint of $AB$,

So the position vector of $D$ is
$\boxed{\vec d=\frac{\vec a+\vec b}{2}}$

(b) Given $A(2,3)$ and $B(4,1)$,

Hence, $\boxed{\vec d=3\vec i+2\vec j}$

(c) Let the position vectors of $A,B,C$ be $\vec a,\vec b,\vec c$.
Since $D$ and $E$ are midpoints of $AB$ and $AC$,

Now,

But,

Therefore,

Hence proved.

(a) Shape formed on the paper

The shape is an ellipse — a closed oval curve.

Reason: The cutting plane makes an angle with the axis of the cone which is greater than the semi-vertical angle $\alpha$ but less than $90^\circ$. Such a plane cuts the cone right round its curved surface, so the boundary of the section is a closed curve, i.e. an ellipse. (If the angle equals $\alpha$ the section is a parabola, and if it is less than $\alpha$ it is a hyperbola.)

(b) Vector proof of Shreya's claim

Let $O$, the centre of the semicircle, be the origin. Let the radius be $r$.

Take $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$, so that $|\vec{a}| = |\vec{b}| = r$.

Since $AOC$ is a diameter, $O$ is the midpoint of $AC$, hence

$\vec{OC} = -\vec{OA} = -\vec{a}.$

Now

$\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a},$

$\vec{CB} = \vec{OB} - \vec{OC} = \vec{b} - (-\vec{a}) = \vec{b} + \vec{a}.$

Taking the dot product,

$\vec{AB}\cdot\vec{CB} = (\vec{b}-\vec{a})\cdot(\vec{b}+\vec{a}) = \vec{b}\cdot\vec{b} - \vec{a}\cdot\vec{a} = |\vec{b}|^2 - |\vec{a}|^2 = r^2 - r^2 = 0.$

Since the dot product of $\vec{AB}$ and $\vec{CB}$ is zero, the two vectors are perpendicular. Therefore the chords $AB$ and $BC$ meet at a right angle, i.e. $\angle ABC = 90^\circ$. Shreya's claim is proved.

(c) Does the claim hold when $B$ lies inside the semicircle?

No. If $B$ lies inside the semicircle, then $|\vec{OB}| = |\vec{b}| < r$. The same working as above gives

$\vec{AB}\cdot\vec{CB} = |\vec{b}|^2 - r^2 < 0.$

A negative dot product means the angle between the two chords is obtuse, not $90^\circ$. So the chords no longer meet at a right angle, and the claim fails. (The right angle occurs only when $B$ lies exactly on the arc, where $|\vec{OB}| = r$.)

(d) Finding the value of $x$

Given: $(AB)^2 = \sqrt{x+9}$, $(BC)^2 = \sqrt{x}$, $AC = 3$ units, and $\angle ABC = 90^\circ$.

Formula: In right-angled $\triangle ABC$ with the right angle at $B$,

$(AB)^2 + (BC)^2 = (AC)^2.$

Substituting:

$\sqrt{x+9} + \sqrt{x} = 3^2$

$\sqrt{x+9} + \sqrt{x} = 9 \quad \text{... (i)}$

Also,

$\left(\sqrt{x+9}\right)^2 - \left(\sqrt{x}\right)^2 = (x+9) - x = 9.$

$\left(\sqrt{x+9} - \sqrt{x}\right)\left(\sqrt{x+9} + \sqrt{x}\right) = 9.$

Using (i):

$\left(\sqrt{x+9} - \sqrt{x}\right) \times 9 = 9$

$\sqrt{x+9} - \sqrt{x} = 1 \quad \text{... (ii)}$

Adding (i) and (ii):

$2\sqrt{x+9} = 10 \;\Rightarrow\; \sqrt{x+9} = 5 \;\Rightarrow\; x + 9 = 25 \;\Rightarrow\; x = 16.$

Check: $\sqrt{16+9} + \sqrt{16} = 5 + 4 = 9 = (AC)^2$. ✓

$\boxed{x = 16}$