Vectors: scalar and vector products and their applications

The projection (scalar component) of $\vec a$ on $\vec b$ is

$\frac{\vec a\cdot\vec b}{|\vec b|}$

Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$, this equals $|\vec a|\cos\theta$, the length of the shadow of $\vec a$ along $\vec b$.

(b) $-2\hat{k}$

Using the determinant form,

(a) For two vectors with angle $\theta$ between them,

$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta,\qquad |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$

Both use the same magnitudes and the same angle, so they are linked:

$\tan\theta=\frac{|\vec a\times\vec b|}{\vec a\cdot\vec b},\qquad (\vec a\cdot\vec b)^2+|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2$

The dot product measures how much the vectors point the same way (zero when perpendicular), while the cross product measures how much they differ in direction (zero when parallel). Where one is large the other is small.

(b) Let $\cos A:\cos B=a:b$, i.e. $\dfrac{\cos A}{\cos B}=\dfrac ab$. By the sine rule $\dfrac ab=\dfrac{\sin A}{\sin B}$, so

$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}\;\Rightarrow\;\sin B\cos A-\cos B\sin A=0\;\Rightarrow\;\sin(B-A)=0$

As $A,B$ are angles of a triangle, $|B-A|<\pi$, so $A=B$. Hence $a=b$ and the triangle is isosceles.

(c) The normal to $y^2=4ax$ at $(at^2,2at)$ has slope $-t$:

$y-2at=-t(x-at^2)\;\Rightarrow\;tx+y-2at-at^3=0$

Compare with $lx+my+n=0$:

$\frac tl=\frac1m=\frac{-(2at+at^3)}{n}$

From the first equality $t=\dfrac lm$. From the second, $n=-m\,t\,a(2+t^2)$. Substituting $t=\dfrac lm$,

$n=-al\left(2+\frac{l^2}{m^2}\right)\;\Rightarrow\;nm^2=-al(2m^2+l^2)$

$\boxed{al^3+2alm^2+nm^2=0}$

(c) $\tan\theta$

Since $|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$ and $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$, the ratio

$\frac{|\vec a \times \vec b|}{\vec a \cdot \vec b} = \frac{|\vec a||\vec b|\sin\theta}{|\vec a||\vec b|\cos\theta} = \tan\theta$

(as $\vec a,\vec b$ are non-zero, $|\vec a||\vec b| \neq 0$ and $\cos\theta \neq 0$).

Let the plane be inclined at angle $\theta$ and the weight be $F$.

Force $A$ parallel to the plane: resolving along the plane,

$A=F\sin\theta\quad\cdots(1)$

Force $B$ parallel to the base (horizontal): its component along the plane is $B\cos\theta$, so

$B\cos\theta=F\sin\theta\;\Rightarrow\;B=F\tan\theta\quad\cdots(2)$

From (1) and (2), $\sin\theta=\dfrac AF$ and $\cos\theta=\dfrac{F\sin\theta}{B}=\dfrac AB$.

Using $\sin^2\theta+\cos^2\theta=1$,

$\frac{A^2}{F^2}+\frac{A^2}{B^2}=1$

Dividing by $A^2$,

$\boxed{\frac1{A^2}=\frac1{B^2}+\frac1{F^2}}$

Hence proved.

Justification:

Let $\vec a$ and $\vec b$ be the adjacent sides of a parallelogram, and let $\theta$ be the angle between them. Then

• base $= |\vec a|$
• height $= |\vec b|\sin\theta$

So,

Thus the magnitude of the vector product $\vec a \times \vec b$ gives the area of the parallelogram whose adjacent sides are $\vec a$ and $\vec b$. The direction of $\vec a \times \vec b$ is perpendicular to the plane of the parallelogram.

Example:

Take adjacent sides

Then

Hence,

Therefore, the area of the parallelogram is $\boxed{\sqrt3}$ square units.

(b) $45^\circ$

Since $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$ and $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$, the given condition gives $\cos\theta=\sin\theta$, so $\tan\theta=1$ and $\theta=45^\circ$.

Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ meeting at $O$, taken as the origin. Let $\vec{OA}=\vec a$ and $\vec{OB}=\vec b$. The diagonals of a parallelogram bisect each other, so $\vec{OC}=-\vec a$ and $\vec{OD}=-\vec b$.

Then $\vec{AB}=\vec b-\vec a,\qquad \vec{BC}=-\vec a-\vec b$

All sides of a rhombus are equal, so $|\vec{AB}|=|\vec{BC}|$:

$|\vec b-\vec a|^2=|\vec a+\vec b|^2$

$|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2\;\Rightarrow\;\vec a\cdot\vec b=0$

Now $\vec{AC}=-2\vec a$ and $\vec{BD}=-2\vec b$, so

$\vec{AC}\cdot\vec{BD}=4\,\vec a\cdot\vec b=0$

Hence the diagonals are perpendicular.

The area of a parallelogram with diagonals $\vec d_1,\vec d_2$ is $\tfrac12|\vec d_1\times\vec d_2|$.

$|\vec d_1\times\vec d_2|=\sqrt{9+25+121}=\sqrt{155}$

$\text{Area}=\frac{\sqrt{155}}{2}\text{ square units}\ (\approx 6.22)$

$\vec{a}-\vec{d}$

From $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$, subtract the second equation from the first: $\vec{a}\times\vec{b}-\vec{a}\times\vec{c}-\vec{c}\times\vec{d}+\vec{b}\times\vec{d}=0$ $\vec{a}\times(\vec{b}-\vec{c})-\vec{c}\times\vec{d}+\vec{b}\times\vec{d}=0$ Using $\vec{c}\times\vec{d}=-\vec{d}\times\vec{c}$ and $\vec{b}\times\vec{d}=-\vec{d}\times\vec{b}$: $\vec{a}\times(\vec{b}-\vec{c})+\vec{d}\times\vec{c}-\vec{d}\times\vec{b}=0$ $\vec{a}\times(\vec{b}-\vec{c})-\vec{d}\times(\vec{b}-\vec{c})=0$ $(\vec{a}-\vec{d})\times(\vec{b}-\vec{c})=0$ Thus $(\vec{a}-\vec{d})$ is parallel to $(\vec{b}-\vec{c})$.

So the correct option is (c) $\vec{a}-\vec{d}$.

100g Newtons

The gravitational force (weight) exerted by the earth on a mass $m$ is $W = mg$. For $m = 100\ \text{kg}$, this gives $W = 100g$ Newtons.

Let like parallel forces $P$ at $A$ and $Q$ at $B$ act with $AB=d$. Their resultant is $R=P+Q$.

Before interchange: the resultant acts at $C$ on $AB$ with $P\cdot AC=Q\cdot CB$, so

$AC=\frac{Qd}{P+Q}$

After interchange ($Q$ at $A$, $P$ at $B$): the resultant acts at $C'$ with $Q\cdot AC'=P\cdot C'B$, so

$AC'=\frac{Pd}{P+Q}$

Displacement:

$CC'=AC'-AC=\frac{Pd-Qd}{P+Q}=\frac{(P-Q)\,d}{P+Q}\qquad(P>Q)$

Hence the point of application of the resultant is displaced by $\dfrac{(P-Q)d}{P+Q}$.

Two points do not fix a unique plane, so the question is read as asking for the direction cosines of the line joining $A(2,0,0)$ and $B(0,2,2)$.

Direction ratios of $AB$: $(0-2,\ 2-0,\ 2-0)=(-2,2,2)\propto(-1,1,1)$, with $\sqrt{1+1+1}=\sqrt3$.

$\text{Direction cosines}=\left(-\frac1{\sqrt3},\ \frac1{\sqrt3},\ \frac1{\sqrt3}\right)$

(Reversing the direction gives $\left(\tfrac1{\sqrt3},-\tfrac1{\sqrt3},-\tfrac1{\sqrt3}\right)$.)

The coordinates of the points and the equation of the line are missing from the question, so the numerical plane cannot be fixed. The method is as follows.

Let the points be $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, and let the line have direction ratios $(l,m,n)$.

1. The plane contains $\overrightarrow{AB}$ and is parallel to the line, so its normal is $\vec N=\overrightarrow{AB}\times(l\hat i+m\hat j+n\hat k)$.
2. If $\vec N=(p,q,r)$, the plane is $p(x-x_1)+q(y-y_1)+r(z-z_1)=0$.

They are perpendicular (orthogonal) to each other.

Since $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0$ and both vectors are non-zero, we get $\cos\theta = 0$, so $\theta = 90^\circ$, i.e. $\vec{a} \perp \vec{b}$.

Two vectors $\vec a$ and $\vec b$ are parallel if one is a scalar multiple of the other, i.e. $\vec a=k\vec b$ for some scalar $k$. Equivalently, $\vec a\times\vec b=\vec 0$, or in components $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\dfrac{a_3}{b_3}$.

Given: $\vec{a} = 2\hat{i} + 3\hat{j}, \qquad \vec{b} = \hat{i} - \hat{j}$

Formula: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2$

Substituting: $\vec{a} \cdot \vec{b} = (2)(1) + (3)(-1)$ $= 2 - 3$ $= -1$

Therefore, $\boxed{\vec{a} \cdot \vec{b} = -1}$

$\vec{AB}=(2-1,\,-1-2,\,4-3)=(1,-3,1),\qquad \vec{AC}=(3-1,\,1-2,\,2-3)=(2,-1,-1)$

$|\vec{AB}\times\vec{AC}|=\sqrt{16+9+25}=\sqrt{50}=5\sqrt2$

$\text{Area}=\tfrac12\times5\sqrt2=\frac{5\sqrt2}{2}\text{ square units}$

(a) $-6\hat{k}$

$\vec b\times\vec a=(3\hat j)\times(2\hat i)=6(\hat j\times\hat i)=6(-\hat k)=-6\hat k$, since $\hat i\times\hat j=\hat k$ and the cross product is anticommutative.

(d) 98 N

Force is weight $=mg=10\times9.8=98$ N. Joule is a unit of energy, not force.

Difference: Like parallel forces act in the same direction along parallel lines, while unlike parallel forces act in opposite directions.

Problem: Let the beam $AB$ have length $2a$, with its weight 60 kg-wt acting at the centre $O$. Strings at $A$ and $B$ carry tensions $T_A$ and $T_B$. Place the 30 kg body at distance $x$ from $O$ towards $A$, so that the string at $A$ is the one that just breaks, $T_A=52.5$.

Vertical balance: $T_A+T_B=60+30=90$, so $T_B=90-52.5=37.5<52.5$ (the other string is safe).

Moments about $B$: $T_A\cdot2a=60\cdot a+30\,(a+x)$ $52.5\times2a=90a+30x\ \Rightarrow\ 105a-90a=30x\ \Rightarrow\ x=\frac a2.$

The body must be placed $\dfrac a2$ from the centre, i.e. one quarter of the beam's length from the centre (towards the string that breaks).

(D) Magnitude of vector product of two vectors along $AB$ and $AD$.

The area of a parallelogram is $|\vec{AB}\times\vec{AD}|$, the vector product of two adjacent sides. In (A) and (C) one vector is a diagonal, and in (B) $AB\parallel DC$ so the product is zero.

(D) parallel

The ceiling and the floor are parallel planes, so the lines of action of $P$ and $Q$ are parallel. Their directions are not specified, so they cannot be called definitely like or unlike, and they lie in different planes, so they are not collinear.

(a) Given $y=\sin^{-1}x$ with $y>0$, we have $\sin y=x$ and $0<y\le\dfrac\pi2$, so $\cos y\ge0$.

$\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$

$\tan y=\frac{\sin y}{\cos y}=\frac{x}{\sqrt{1-x^2}}$

(b) Given $\vec a\times\vec b=\vec a\times\vec c$ with $\vec a\ne\vec 0$:

$\vec a\times(\vec b-\vec c)=\vec 0$

so $\vec b-\vec c$ is parallel to $\vec a$ (or zero), i.e. $\vec b-\vec c=\lambda\vec a$.

The cross-product condition alone does not force $\lambda=0$: for $\vec a=(1,0,0)$, $\vec b=(0,1,0)$, $\vec c=(2,1,0)$ both products equal $(0,0,1)$ but $\vec b\ne\vec c$. The result $\vec b=\vec c$ needs the companion condition $\vec a\cdot\vec b=\vec a\cdot\vec c$ (usually stated with this problem). Then $\vec a\cdot(\vec b-\vec c)=\lambda|\vec a|^2=0$, so $\lambda=0$ because $\vec a\neq\vec 0$, and therefore $\vec b=\vec c$.

Let the particle be projected with speed $v$ at angle $\theta$ to the horizontal. Then

$H=\frac{v^2\sin^2\theta}{2g},\qquad R=\frac{v^2\sin2\theta}{g}=\frac{2v^2\sin\theta\cos\theta}{g}$

From the first, $\dfrac{v^2\sin^2\theta}{g}=2H$, and so $\dfrac{v^2\cos^2\theta}{g}=\dfrac{v^2}{g}-2H$. Then

$R^2=4\left(\frac{v^2\sin^2\theta}{g}\right)\left(\frac{v^2\cos^2\theta}{g}\right)=8H\left(\frac{v^2}{g}-2H\right)$

$R=4\sqrt{H\left(\frac{v^2}{2g}-H\right)}$

Check: at $\theta=45^\circ$, $H=\dfrac{v^2}{4g}$ and $R=\dfrac{v^2}{g}$, which agrees with this formula. The form printed in the question, $R=\sqrt{H\left(\frac{2v^2}{g}-H\right)}$, does not hold (it gives a different value at $45^\circ$), so the question appears to be mistyped; the correct relation is the one derived above.

(a) For $P(4,4,-10)$ and $Q(-2,2,4)$, the direction ratios of $PQ$ are $(-6,-2,14)$, with magnitude $\sqrt{36+4+196}=\sqrt{236}=2\sqrt{59}$.

$\text{Direction cosines}=\left(-\frac{3}{\sqrt{59}},\ -\frac{1}{\sqrt{59}},\ \frac{7}{\sqrt{59}}\right)$

(the signs reverse for the opposite direction $QP$).

(b) Take a cube of side $a$ with one vertex at the origin. Two body diagonals are $OA$ from $(0,0,0)$ to $(a,a,a)$ and $BC$ from $(a,0,0)$ to $(0,a,a)$, with direction vectors $\vec d_1=(a,a,a)$ and $\vec d_2=(-a,a,a)$.

$\cos\theta=\frac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|}=\frac{-a^2+a^2+a^2}{(a\sqrt3)(a\sqrt3)}=\frac13$

$\theta=\cos^{-1}\frac13\approx70^\circ32'$

(c) Dividing $16(y-1)^2-9(x-5)^2=144$ by $144$,

$\frac{(y-1)^2}{9}-\frac{(x-5)^2}{16}=1$

This is a hyperbola with centre $(5,1)$ and vertical transverse axis, with $a^2=9$, $a=3$. The vertices are $(5,1\pm3)$, i.e. $(5,4)$ and $(5,-2)$.