Question

Let two non-collinear unit vectors \(\hat{a}\) and \(\hat{b}\) form an acute angle. A point \(P\) moves so that at any time \(t\) the position vector \(\overrightarrow{O P}\) (where \(O\) is the origin) is given by \(\hat{a} \cos t+\hat{b} \sin t .\) When \(P\) is farthest from origin \(O\), let \(M\) be the length of \(O P\) and \(\hat{u}\) be the unit vector along \(\overrightarrow{O P}\). Then, (A) \(\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) and \(M=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}\) (B) \(\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}\) and \(M=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}\) (C) \(\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) and \(M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}\) (D) \(\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}\) and \(M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}\)

Short answer

(A) \(\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) and \(M=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}\)

Step-by-step solution

Find the Position Vector

Write the position vector of the point P in terms of given unit vectors and trigonometric functions: \(\overrightarrow{OP} = \hat{a} \cos t + \hat{b} \sin t\)

Find the Length of the Position Vector

To find the maximum distance (M), compute the length (magnitude) of the position vector: \(M = |\overrightarrow{OP}| = |\hat{a} \cos t + \hat{b} \sin t|\).

Calculate the Magnitude Square of OP

Express the magnitude square of the position vector using the dot product: \(M^2 = (\hat{a} \cos t + \hat{b} \sin t) \cdot (\hat{a} \cos t + \hat{b} \sin t)\)

Simplify the Magnitude Square

Expand the dot product to simplify the expression: \(M^2 = \hat{a} \cdot \hat{a} \cos^2 t + 2\hat{a} \cdot \hat{b} \sin t\cos t + \hat{b} \cdot \hat{b} \sin^2 t\) considering \(\hat{a}\) and \(\hat{b}\) to be unit vectors.

Maximize M Using Trigonometry

To maximize M, use the trigonometric identity \(\sin t\cos t = \frac{1}{2}\sin 2t\) and that \(\sin^2 2t\) will be maximum when \(\sin 2t = 1\).

Express the Maximum Length

Rewrite M in terms of maximum point which makes the term with \(\sin 2t\) maximum: \(M^2 = \cos^2 t + \sin^2 t + 2\hat{a} \cdot \hat{b} \sin t\cos t\) gives us \(M = (1 + 2\hat{a}\cdot \hat{b})^{\frac{1}{2}}\) when \(\sin 2t = 1\).

Determine the Unit Vector Along OP

Find the unit vector \(\hat{u}\) along \(\overrightarrow{OP}\) when P is farthest from the origin (when \(\sin t = \cos t = \frac{1}{\sqrt{2}}\)): \(\hat{u} = \frac{\overrightarrow{OP}}{|\overrightarrow{OP}|} = \frac{\hat{a} \frac{1}{\sqrt{2}} + \hat{b} \frac{1}{\sqrt{2}}}{M}\).

Simplify the Unit Vector Expression

Simplify \(\hat{u}\) to get \(\hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}\) as \(\frac{1}{\sqrt{2}}\) will cancel out with part of M's expression.

Conclusion

Conclude that the correct expressions for M and \(\hat{u}\) are (A) \(\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) and \(M=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}\), because M is derived from the maximum of the original magnitude. Thus, answer (A) is correct.

Key concepts

Position Vector

Imagine standing at a point in space—it could be anywhere—and pointing directly to an object. The arrow you mentally draw from your position to the object is the essence of what we call a position vector. It serves as a directional line from a fixed reference point, often termed the origin, to any point in space. In the context of JEE Advanced Physics, understanding position vectors is crucial since they lay the groundwork for defining motion and forces in vectorial terms.

In a two-dimensional or three-dimensional space, a position vector not only tells the direction toward the point of interest, such as point P in our exercise, but also the distance from the origin to that point. Mathematically, it's depicted as \(\overrightarrow{OP}\) for a point P from the origin O. A position vector can be expressed in terms of its components along the base vectors, which, in this case, are given as the unit vectors \(\hat{a}\) and \(\hat{b}\) along two non-collinear directions.

Vector Magnitude Calculation

The magnitude of a vector represents its 'length' or 'size' and is a measure of how far the end point is from the start. In the realm of JEE Advanced Physics, calculating the magnitude of vectors is pivotal for grasping concepts that range from forces to field strengths.

To calculate the vector magnitude, one might use the Pythagorean theorem in two dimensions, or its extension, in three dimensions. If you know the vector’s components along the base axes, the magnitude can be found by taking the square root of the sum of the squares of these components. For a vector expressed as \(\vec{A} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}\), its magnitude is \(|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\). Understanding this formula is essential for students to analyze vector-related problems effectively.

Trigonometric Identity in Vectors

Trigonometry and vectors intertwine beautifully in physics problems, where angles and magnitudes come into play. One common intersection of these two fields is the usage of trigonometric identities to simplify vector calculations.

For instance, in our JEE problem, the trigonometric identity \(\sin t\cos t = \frac{1}{2}\sin 2t\) converts what would be a complex computation into a more manageable form. This is because many vector magnitudes involve the dot product (which we'll explore next), and can thus be expressed in terms of sine and cosine functions of angles. The manipulation of these functions through trigonometric identities often leads to simpler expressions that are easier to maximize or minimize, a common task in physics problem-solving.

Dot Product of Vectors

Venturing into a core aspect of vector algebra, the dot product (or scalar product) is a method to multiply two vectors together and obtain a scalar quantity. This scalar reveals how much one vector extends in the direction of another—their 'overlap', so to speak. The formula for the dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos\theta\), where \(\theta\) is the angle between them.

In our JEE problem, the dot product is utilized to find the square of the position vector’s magnitude. This concept is invaluable for understanding the projections of vectors, energy calculations in physics, and much more. Importantly, the dot product's properties often simplify to common trigonometric identities when vectors are unit vectors, which is a handy technique for solving many physics problems.