Fundamentals of Vectors Questions in English

(C) Let the magnitude of the two vectors be $A$ and $B$. Given that they have the same magnitude,let $A = B = x$.
The resultant magnitude $R$ is given by the formula $R^2 = A^2 + B^2 + 2AB \cos \theta$,where $\theta$ is the angle between the vectors.
According to the problem,twice the product of the magnitudes is equal to the square of the resultant: $2(AB) = R^2$.
Substituting $A = x$ and $B = x$,we get $2(x \cdot x) = x^2 + x^2 + 2(x \cdot x) \cos \theta$.
This simplifies to $2x^2 = 2x^2 + 2x^2 \cos \theta$.
Subtracting $2x^2$ from both sides,we get $0 = 2x^2 \cos \theta$.
Since $x \neq 0$,we must have $\cos \theta = 0$.
Therefore,$\theta = 90^{\circ}$.

(A) The force vector is given by $\vec{F} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.
Let the horizontal direction be represented by the unit vector $\hat{u}$ along $\hat{i} + \hat{j}$.
$\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
The angle $\theta$ between the force $\vec{F}$ and the horizontal direction $\hat{u}$ is given by $\cos \theta = \frac{\vec{F} \cdot \hat{u}}{|\vec{F}| |\hat{u}|}$.
First,calculate the magnitude of $\vec{F}$: $|\vec{F}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.
Next,calculate the dot product $\vec{F} \cdot \hat{u} = (4\hat{i} + 3\hat{j} - 5\hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{4 + 3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$.
Thus,$\cos \theta = \frac{7/\sqrt{2}}{5\sqrt{2}} = \frac{7}{10}$.
Note: Based on the provided options,the intended calculation assumes the horizontal component is simply the $x$-component projection. If we define the horizontal direction as the $x$-axis $(\hat{i})$,then $\cos \theta = \frac{F_x}{|F|} = \frac{4}{5\sqrt{2}} = \frac{2\sqrt{2}}{5}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2\sqrt{2}}{5}\right)$.

(C) Let the resultant be $\vec{R} = \vec{A} + \vec{B}$.
Given that $\vec{R} \perp \vec{A}$,the dot product $\vec{A} \cdot \vec{R} = 0$.
$\vec{A} \cdot (\vec{A} + \vec{B}) = 0 \implies A^2 + AB \cos \theta = 0 \implies AB \cos \theta = -A^2$ ... $(i)$
Given the magnitude $R = \frac{B}{2}$,we have $R^2 = \frac{B^2}{4}$.
Using the law of vector addition,$R^2 = A^2 + B^2 + 2AB \cos \theta$.
Substituting $AB \cos \theta = -A^2$ into the equation:
$\frac{B^2}{4} = A^2 + B^2 + 2(-A^2) = B^2 - A^2$.
Rearranging gives $A^2 = B^2 - \frac{B^2}{4} = \frac{3B^2}{4}$,so $A = \frac{\sqrt{3}}{2}B$.
Substituting $A$ back into $(i)$:
$B(\frac{\sqrt{3}}{2}B) \cos \theta = -(\frac{\sqrt{3}}{2}B)^2$.
$\frac{\sqrt{3}}{2} B^2 \cos \theta = -\frac{3}{4} B^2$.
$\cos \theta = -\frac{3}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{\sqrt{3}}{2}$.
Thus,$\theta = 150^{\circ}$.

(D) The magnitude of the resultant $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given that the magnitudes of the two vectors are equal,let $A = B = x$.
It is also given that the magnitude of the resultant is equal to the magnitude of either vector,so $R = x$.
Substituting these values into the formula: $x = \sqrt{x^2 + x^2 + 2x^2 \cos \theta}$.
Squaring both sides: $x^2 = 2x^2 + 2x^2 \cos \theta$.
Dividing by $x^2$ (assuming $x \neq 0$): $1 = 2 + 2 \cos \theta$.
Rearranging the terms: $2 \cos \theta = 1 - 2 = -1$.
Therefore,$\cos \theta = -\frac{1}{2}$.
This corresponds to an angle of $\theta = 120^{\circ}$.

(B) Given vector $P = 3 \hat{i} + 4 \hat{j}$.
Let the direction vector be $Q = \hat{i} + 2 \hat{j}$.
The component of vector $P$ along the direction of vector $Q$ is given by the projection formula: $P_{Q} = \frac{P \cdot Q}{|Q|}$.
First,calculate the dot product $P \cdot Q = (3 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = (3 \times 1) + (4 \times 2) = 3 + 8 = 11$.
Next,calculate the magnitude of vector $Q$: $|Q| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.
Therefore,the component is $\frac{P \cdot Q}{|Q|} = \frac{11}{\sqrt{5}}$.

(B) For any vector,the direction cosines are defined as $\cos \alpha, \cos \beta$ and $\cos \gamma$.
We know the fundamental identity for direction cosines is $\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma = 1$.
Using the trigonometric identity $\sin ^2 \theta = 1 - \cos ^2 \theta$,we can rewrite the expression as:
$(1 - \sin ^2 \alpha) + (1 - \sin ^2 \beta) + (1 - \sin ^2 \gamma) = 1$.
$3 - (\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma) = 1$.
$\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma = 2$.
Therefore,$\sin ^2 \alpha + \sin ^2 \beta = 2 - \sin ^2 \gamma$.
Since $\sin ^2 \gamma = 1 - \cos ^2 \gamma$,we substitute this into the equation:
$\sin ^2 \alpha + \sin ^2 \beta = 2 - (1 - \cos ^2 \gamma) = 1 + \cos ^2 \gamma$.

(B) We know that the magnitude of the resultant $R$ of two vectors $A$ and $B$ is given by the formula:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Here,$R$ is maximum when $\cos \theta$ is maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0^{\circ}$.
Substituting $\theta = 0^{\circ}$ into the formula:
$R_{\max} = \sqrt{A^2 + B^2 + 2AB(1)} = \sqrt{(A+B)^2} = A + B$.
Therefore,the resultant is maximum when the angle between the two vectors is $0^{\circ}$.

(A) The position vector of point $A$ is $\vec{r}_A = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$.
The position vector of point $B$ is $\vec{r}_B = 6 \hat{i} + 6 \hat{j} + 9 \hat{k}$.
The displacement vector $\vec{d}$ is given by the change in position: $\vec{d} = \vec{r}_B - \vec{r}_A$.
$\vec{d} = (6 - 2) \hat{i} + (6 - 2) \hat{j} + (9 - 3) \hat{k}$.
$\vec{d} = 4 \hat{i} + 4 \hat{j} + 6 \hat{k}$.

(D) vector $\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = 1$.
Given the vector is $0.5 \hat{i} + 0.8 \hat{j} + c \hat{k}$.
Therefore,$\sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1$.
Squaring both sides,we get $(0.5)^2 + (0.8)^2 + c^2 = 1^2$.
$0.25 + 0.64 + c^2 = 1$.
$0.89 + c^2 = 1$.
$c^2 = 1 - 0.89 = 0.11$.
Thus,$c = \sqrt{0.11}$.