Q.If 𝑎⃗ = 3𝑖̂ + 2𝑗̂ + 4𝑘̂ , 𝑏⃗⃗ = 𝑖̂ + 𝑗̂ − 3𝑘̂ and 𝑐⃗ = 6𝑖̂ − 𝑗̂ + 2𝑘̂ are three given vectors, then (2𝑎⃗. 𝑖̂)𝑖̂ − (𝑏⃗⃗. 𝑗̂)𝑗̂ + (𝑐⃗. 𝑘̂)𝑘̂ is same as the vector
(A) 𝑎⃗
(B) 𝑏⃗⃗ + 𝑐⃗
(C) 𝑎⃗ − 𝑏⃗⃗
(D) 𝑐⃗

A unit vector along the X-axis is $\hat\imath = (1,0,0)$. We require

Therefore …

Concept: Vector addition; a unit vector along the X-axis is i.

First add the two given vectors:

(i - 3j + 2k) + (3i + 6j - 7k) = (1+3)i + (-3+6)j + (2-7)k = 4i + 3j - 5k

Let the required vector be v. Then

(4i + 3j - 5k) + v = i

v = i - (4i + 3j - 5k) = (1 - 4)i + (0 - 3)j + (0 + 5)k

v = -3i - 3j + 5k …

Concept: resolve the two forces; the resultant is perpendicular to one of them, so the component of the resultant along that force is zero.

Let P lie along the x-axis and Q make 120 degrees with it.

Q components: (-Q/2, +Q sqrt(3)/2)

Resultant components: (P - Q/2, Q sqrt(3)/2)

Resultant perpendicular to P => x-component = 0:

P - Q/2 = 0 -> Q = 2P

Magnitude of the resultant: …

Concept and Intuition

When two forces act at a point, their resultant's magnitude and direction are given by the parallelogram law. Here, we know the resultant is perpendicular to the smaller force. That means the smaller force has no component along the resultant's direction — it is entirely "sideways" to the resultant. This geometric condition gives a clean relationship between the two forces and the resultant, which, together with the sum of their magnitudes, lets us solve for each.

Step-by-step solution

1. Set up variables and given conditions Let the two forces be $A$ and $B$, with $A > B$ (so $B$ is the smaller force). Given:

Resultant magnitude $R = 20$ N.

The resultant is at $90^\circ$ to the smaller force $B$.

2. Apply the parallelogram law For two forces $A$ and $B$ with an angle $\theta$ between them, the resultant magnitude is:

Also, the angle $\alpha$ that the resultant makes with force $B$ satisfies:

Here, $\alpha = 90^\circ$ (resultant perpendicular to $B$), so $\tan 90^\circ$ is undefined, meaning the denominator must be zero:

3. Substitute $\cos\theta$ into the resultant equation From $R^2 = A^2 + B^2 + 2AB\cos\theta$, replace $\cos\theta$:

Simplify:

So:

4. Solve the system of equations From (1): $A = 50 - B$. Substitute into (2):

Expand:

Then $A = 50 - 21 = 29$.

5. Check the "smaller force" condition We assumed $B$ is the smaller force, but here $B = 21$ and $A = 29$ — that's fine, $B < A$. However, does the resultant being perpendicular to the smaller force hold? Yes, because we used that condition. But wait — let's verify the resultant magnitude:

So $R = \sqrt{400} = 20$ N. This matches option (C): 29 N, 21 N. …

Step 1 — The concept: angular velocity is an axial vector.

Unlike a linear velocity, $\vec\omega$ does not point along the direction anything is moving. It points along the axis of rotation, and its sense is given by the right-hand rule: curl the fingers of the right hand in the sense of the rotation, and the extended thumb gives the direction of $\vec\omega$. Its magnitude is $\omega = d\theta/dt$.

Step 2 — Fix the axis.

The fan hangs from a vertical down-rod and spins about that rod. The dashed axle is vertical, and in the given frame $Z$ points straight up. So $\vec\omega$ must be along $+Z$ or $-Z$ — the horizontal directions $X$ and $Y$ are impossible, immediately eliminating options (A) and (B). The only question left is the sign.

Step 3 — Read the sense of rotation from the figure.

The frame drawn is right-handed: $Z$ up, $Y$ to the right, $X$ out of the page towards the viewer. The rotation arrow is an ellipse round the axle whose near (front, $+X$ side) arc runs from right to LEFT, while the far arc runs left to right.

So a blade tip that is momentarily nearest the viewer (position $\vec r$ along $+\hat X$) has velocity $\vec v$ pointing in the $-\hat Y$ direction.

Step 4 — Apply $\vec v = \vec\omega \times \vec r$ to pin the sign.

Write $\vec\omega = \omega_z \hat Z$ and take the blade tip at $\vec r = r\hat X$:

But the figure says this point moves along $-\hat Y$, i.e. $\vec v = -|v|\hat Y$. Matching:

$\omega_z r = -|v| \;\Longrightarrow\; \omega_z < 0 .$ …

With $\vec p=5\hat i-4\hat j+3\hat k$ and $\vec q=3\hat i+2\hat j-\hat k$:

$\hat i:\;(-4)(-1)-(3)(2)=4-6=-2$

$\hat j:\;-[(5)(-1)-(3)(3)]=-[-5-9]=14$

$\hat k:\;(5)(2)-(-4)(3)=10+12=22$ …