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Q.The unit vector along (i + 2j + k) will be:

(a) i + 2j + k
(b) (i + 2j - k)/√6
(c) (i + 2j - k)/6
(d) (i + 2j - k)/√3
Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2024MCQ· 1mImportance★★★★★
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Concept understanding — Vector Component Extraction

Vector Component Extraction: The Intuition

Imagine pushing a heavy box across the floor at an angle — not straight forward, but slightly downward. Some of your effort moves the box forward, and some presses it into the floor. The force you apply is a single vector, but its effect splits into two independent directions: horizontal and vertical.

That splitting is vector component extraction. Any vector can be seen as the sum of two (or three) simpler vectors pointing along chosen reference directions — usually the coordinate axes. Each of those simpler vectors is a component.

Note

"Component" means "a part of a whole." In vectors, the components are the parts that add up to give the original vector.

The Precise Statement

Given a vector v⃗\vec{v} in a plane, and perpendicular axes xx and yy, the components of v⃗\vec{v} are its projections onto those axes:

v⃗=vxi^+vyj^\vec{v} = v_x \hat{i} + v_y \hat{j}

where i^\hat{i} and j^\hat{j} are unit vectors along the xx and yy axes, and vxv_x, vyv_y are scalar components (numbers, possibly negative).

If v⃗\vec{v} makes an angle θ\theta from the positive xx-axis, then:

vx=∣v⃗∣cos⁡θandvy=∣v⃗∣sin⁡θv_x = |\vec{v}| \cos \theta \quad \text{and} \quad v_y = |\vec{v}| \sin \theta

Component along an axis=(magnitude of vector)×cos⁡(angle between vector and that axis)\text{Component along an axis} = (\text{magnitude of vector}) \times \cos(\text{angle between vector and that axis})

Why This Works: The Geometry

Draw a vector from the origin. Drop a perpendicular from its tip to the xx-axis — that gives vxv_x. Drop another to the yy-axis — that gives vyv_y. The original vector is the diagonal of the rectangle formed by vxv_x and vyv_y. This is the Pythagorean theorem in reverse: if you know the hypotenuse and one angle, trigonometry gives you the legs.

A Concrete Example

A force of 10 N acts at 30∘30^\circ above the horizontal.

  • Fx=10cos⁡30∘=10×32=53≈8.66F_x = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66 N
  • Fy=10sin⁡30∘=10×12=5F_y = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5 N

So the force vector is 8.66i^+5j^8.66\hat{i} + 5\hat{j} N.

Watch out

A common mistake: using sin⁡\sin for the horizontal component and cos⁡\cos for the vertical. Check: if the angle is measured from the xx-axis, the side adjacent to it is along xx — that's cos⁡\cos; the opposite side is along yy — that's sin⁡\sin.

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