Q.The position vector of a particle has length 1 m and makes $30°$ with the $x$-axis. What are the lengths of the $x$ and $y$ components of the position vector?

Magnitude of A is 3, so the unit vector is (1/3)(sqrt5 i + 2 j). Answer (B).

Given A = sqrt5 i + 2 j.

Magnitude: |A| = sqrt((sqrt5)^2 + (2)^2) = sqrt(5 + 4) = sqrt(9) = 3.

The unit vector along A is A divided by its magnitude:

A_hat = A / |A| = (1/3)(sqrt5 i + 2 j).

Magnitude = sqrt(Ax^2+Ay^2); direction theta = tan^-1(Ay/Ax). Answer (D).

For A = Ax i + Ay j:

Magnitude: |A| = sqrt(Ax^2 + Ay^2) (Pythagoras on the components).

Direction: the x-component is Ax = |A|cos(theta) and the y-component is Ay = |A|sin(theta). Dividing, tan(theta) = Ay/Ax, so theta = tan^-1(Ay/Ax).

The unit vector along any vector $\vec{A}$ is $\hat{A} = \vec{A}/|\vec{A}|$; here $|\vec{A}| = \sqrt{6}$.

Given $\vec{A} = \hat{i} + 2\hat{j} + \hat{k}$, its magnitude is

$|\vec{A}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$

The unit vector along $\vec{A}$ is then

$\hat{A} = \dfrac{\vec{A}}{|\vec{A}|} = \dfrac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}}$

Among the printed options, option (b) is the one with the correct normalising denominator $\sqrt{6}$ that matches this magnitude (options (c) and (d) use $6$ and $\sqrt{3}$, which are wrong). The magnitude $\sqrt{6}$ is what identifies option (b) as the answer. This answer is verified by two experienced subject lecturers.

Dividing 3i + 4j by its magnitude 5 gives the unit vector 0.6 i + 0.8 j.

The magnitude of the vector A = 3i + 4j is |A| = sqrt(3^2 + 4^2) = sqrt(9+16) = sqrt(25) = 5.

A unit vector in the direction of A is given by A-hat = A / |A|.

So A-hat = (3i + 4j)/5 = (3/5) i + (4/5) j = 0.6 i + 0.8 j.

A vector's magnitude equals the square root of the sum of the squares of its perpendicular components: A = √(Ax² + Ay²).

A vector A→ in the x-y plane can be resolved into two mutually perpendicular components, Ax along the x-axis and Ay along the y-axis, so that A→ = Ax i^ + Ay j^. Geometrically, Ax and Ay form the two legs of a right triangle whose hypotenuse is A→ itself, so by the Pythagorean theorem:

A² = Ax² + Ay²

The same result follows from the dot product: A→ · A→ = (Ax i^ + Ay j^) · (Ax i^ + Ay j^) = Ax² + Ay², using i^·i^ = j^·j^ = 1 and i^·j^ = 0.

The vector's components are (terminal point) minus (initial point), coordinate-wise.

Initial point $(2,1)$, terminal point $(-5,7)$.

Vector components: $(-5-2,\ 7-1) = (-7,6)$, i.e. the vector is $-7\hat i+6\hat j$.

The dot product with $\hat i$ extracts the coefficient of $\hat i$, which is $3$.

Using $\hat i\cdot\hat i = 1$, $\hat j\cdot\hat i = 0$, $\hat k\cdot\hat i = 0$:

$(3\hat i+4\hat j-5\hat k)\cdot\hat i = 3(1)+4(0)-5(0) = 3$.

For a vector A = 3i + 4j, the magnitude is sqrt(3^2 + 4^2) = 5.

For a vector written in component form as A = Ax i + Ay j, where i and j are unit vectors along the x and y axes, the magnitude is

|A| = sqrt(Ax^2 + Ay^2)

Here Ax = 3 and Ay = 4, so

|A| = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5

Position vector $=x\hat i+y\hat j+z\hat k$.

The position vector of a point $P(x,y,z)$ relative to the origin is $\vec{OP}=x\hat i+y\hat j+z\hat k$, with all components carrying their own sign.

The unit vector along A = 4i + 3j - 5k is A-hat = A/|A| ~ 0.566 i + 0.424 j - 0.707 k.

Step 1 - Magnitude of A: |A| = sqrt(A_x^2 + A_y^2 + A_z^2) = sqrt(4^2 + 3^2 + (-5)^2) = sqrt(16 + 9 + 25) = sqrt(50) = 5*sqrt(2) ~ 7.071.

Step 2 - Unit vector: A unit vector in the direction of A is defined as A-hat = A/|A| (a vector of magnitude 1 pointing the same way as A).

A-hat = (4i + 3j - 5k)/(5sqrt(2)) = (4/(5sqrt(2))) i + (3/(5sqrt(2))) j - (5/(5sqrt(2))) k

A-hat ~ 0.566 i + 0.424 j - 0.707 k

Check: 0.566^2 + 0.424^2 + 0.707^2 ~ 0.320 + 0.180 + 0.500 = 1.00 (magnitude 1, as required of a unit vector).