\hat{j} \times \hat{i} =

24:T2311,

The cross product is anticommutative, so reversing the order reverses the direction. From the unit vector rules $\hat{i} \times \hat{j} = \hat{k}$ , we get $\hat{j} \times \hat{i} = - \hat{i} \times \hat{j} = - \hat{k}$ .

Option C is $\hat{i} \times \hat{j}$ , the product in the opposite order. Option A would be correct only for parallel vectors, and $\hat{j}$ and $\hat{i}$ are perpendicular, so their cross product is not zero. Option B is a scalar, but a cross product is a vector.

25:T1267,

The x-component is the one along the horizontal, so $F_{x} = F cos θ$ (equation 2.4): $F_{x} = 10 cos 3 0^{\circ} = 10 \times 0.866 \approx 8.7 N$ .

Option B is $F sin 3 0^{\circ}$ , which is the vertical (y) component, so it is the trap for students who swap sine and cosine.

26:T12b1,

The forces are perpendicular, so they act as the two rectangular components of the resultant. By Pythagoras (equation 2.7): $R = 3^{2} + 4^{2} = 25 = 5 N$ .

Option A is the plain sum, which is correct only for forces in the same direction. Option C is the difference, which is correct only for opposite directions. Option D needs equal and opposite forces.

27:T9e7,

The dot product gives $cos θ = \frac{A \cdot B}{A B}$ (equation 2.14). Taking the inverse cosine gives $θ$ directly from the two magnitudes and the scalar product.

The cross product also contains the angle (through $sin θ$ ), but it gives a vector and the dot product is the standard method. The head to tail rule and the right hand rule are used to find a resultant or a direction, not an angle.

28:T4a9,For which angle the equation

∣ A \cdot B ∣ = ∣ A \times B ∣

is correct?29:T17d3,

Substitute the two products: $A B ∣ cos θ ∣ = A B ∣ sin θ ∣$ . Cancel $A B$ to get $∣ cos θ ∣ = ∣ sin θ ∣$ , so $tan θ = 1$ and $θ = 4 5^{\circ}$ .

At $9 0^{\circ}$ the dot product is zero while the cross product is $A B$ , so option D fails. At $3 0^{\circ}$ and $6 0^{\circ}$ the sine and cosine differ.

2a:T814,

\hat{j} . (\hat{i} \times \hat{k}) =

2b:T4760,

By the unit vector rules $\hat{k} \times \overset{}{^} = \overset{}{^}$ , so reversing the order gives $\overset{}{^} \times \hat{k} = - \overset{}{^}$ . Then $\overset{}{^} \cdot (\overset{}{^} \times \hat{k}) = \overset{}{^} \cdot (- \overset{}{^}) = - (\overset{}{^} \cdot \overset{}{^}) = - 1$ . So B is correct.

Option C ( $1$ ) comes from taking $\overset{}{^} \times \hat{k} = + \overset{}{^}$ , which is true only for the cyclic order $\overset{}{^} \to \overset{}{^} \to \hat{k} \to \overset{}{^}$ . Option A ( $0$ ) would hold if the bracket were perpendicular to $\overset{}{^}$ , but $\overset{}{^} \times \hat{k}$ lies along $\overset{}{^}$ .

2c:T202e,If

A \cdot B = ∣ A \times B ∣

then angle '

θ

' between vectors

A

and

B

is:2d:T2d02,

Put $A \cdot B = A B cos θ$ and $∣ A \times B ∣ = A B sin θ$ in the given condition: $A B cos θ = A B sin θ$ . Cancel $A B$ to get $tan θ = 1$ , so $θ = π /4$ ( $4 5^{\circ}$ ). So B is correct.

At $θ = 0$ or $π$ the cross product is zero while the dot product is $\pm A B$ , and at $θ = π /2$ the dot product is zero while the cross product is $A B$ . So A, C and D cannot satisfy the condition.