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1. Vector Form of Velocity and Acceleration

A vector in 3-D is written in unit-vector form as
$\vec v = v_x\,\hat i + v_y\,\hat j + v_z\,\hat k$
$\vec a = a_x\,\hat i + a_y\,\hat j + a_z\,\hat k$

2. Dot (Scalar) Product

For two vectors $\vec A$ and $\vec B$,
$\vec A \cdot \vec B = |\vec A|\,|\vec B|\,\cos\theta$
where $\theta$ is the angle between them.

• If $\theta = 90^{\circ}$ (right angle) then $\cos\theta = 0$, so
  $\vec A \cdot \vec B = 0.$
• If $\theta = 45^{\circ}$ then $\cos\theta = \tfrac{1}{\sqrt 2}.$

3. Applying to the Given Vectors

Given in the problem:

Velocity:
$\vec v = 2\,\hat i + 3\,\hat j - 4\,\hat k$

Acceleration:
$\vec a = n\,\hat i + 2\,\hat j + 1\,\hat k$

Case in the paper (standard JEE style)

Most books use the fact that velocity and acceleration are perpendicular when the question says "angle between them is 90°" (often printed as a right-angle sign). When that is the case:

$\vec v\cdot\vec a = 0$

4. Solving the Dot Product

$\vec v\cdot\vec a = (2)(n) + (3)(2) + (-4)(1) = 2n + 6 - 4 = 2n + 2$

Set it to zero for $90^{\circ}$:

$2n + 2 = 0 \;\;\Rightarrow\;\; n = -1$

That single algebraic step gives the required value of n.

If, by chance, your sheet actually says the angle is $45^{\circ}$, you would equate the dot product to $|\vec v|\,|\vec a| \cos45^{\circ}$ and you would notice the math leads to no real solution. That usually means there was a misprint – JEE exams deliberately set vectors at 0° or 90° to keep arithmetic clean.

Think of Vectors as Arrows

Imagine you have two arrows stuck in a ball – one arrow shows how fast and in what direction the ball is moving (velocity v) and the other arrow shows how its speed is changing (acceleration a).
If the two arrows are at right-angles to each other, they make a perfect “L” shape – just like the corner of a room.
Mathematically, when two arrows are at 90° the special ‘dot’ calculation we do for them becomes zero.
So, to find the missing number n in the acceleration arrow, we simply make that dot calculation zero and solve.
That’s all!

Step-by-Step Solution

1. Write the vectors
   $\vec v = 2\,\hat i + 3\,\hat j - 4\,\hat k$
   $\vec a = n\,\hat i + 2\,\hat j + 1\,\hat k$

2. Use the perpendicular condition
   $\vec v \cdot \vec a = 0$

3. Compute the dot product
   $\vec v\cdot\vec a = 2n + 6 - 4 = 2n + 2$

4. Set equal to zero and solve for $n$
   $2n + 2 = 0 \;\;\Rightarrow\;\; n = -1$

5. Answer
   $n = -1$