Question

Let \(\triangle P Q R\) be a triangle. Let \(\vec{a}=\overrightarrow{Q R}, \vec{b}=\overrightarrow{R P}\) and \(\vec{c}=\overrightarrow{P Q}\). If \(|\vec{a}|=12,|\vec{b}|=4 \sqrt{3}\) and \(\vec{b}, \vec{c}=24\), then which of the following is (are) true? (A) \(\frac{|\vec{c}|^{2}}{2}-|\vec{a}|=12\) (B) \(\frac{|\vec{c}|^{2}}{2}+|\vec{a}|=30\) (C) \(|\vec{a} \times \vec{b}+\vec{c} \times \vec{a}|=48 \sqrt{3}\) (D) \(\vec{a} \cdot \vec{b}=-72\)

Short answer

Option A is false, as \(\frac{120}{2} - 12 = 48\). Option B is true, since \(\frac{120}{2} + 12 = 72\). Option C requires more information to evaluate. Option D is false as \(\vec{a} \cdot \vec{b}\) equals \(12 \cdot 4\sqrt{3} \cdot (-1)\), which is not equal to -72.

Step-by-step solution

Use the Law of Cosines to find the dot product

Use the Law of Cosines in terms of vectors to find \(\vec{b} \cdot \vec{c}\). The Law of Cosines states that \(\vec{b} \cdot \vec{c} = |\vec{b}|^2 + |\vec{c}|^2 - |\vec{a}|^2\), since \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\). We are given \(\vec{b} \cdot \vec{c} = 24, |\vec{b}| = 4\sqrt{3}\), and \( |\vec{a}| = 12 \). Substitute these values into the Law of Cosines and solve for \( |\vec{c}|^2\).

Calculate \( |\vec{c}|^2 \)

Substituting the given values, we get \(24 = (4\sqrt{3})^2 + |\vec{c}|^2 - 12^2\). This simplifies to \(24 = 48 + |\vec{c}|^2 - 144\), which further reduces to \(|\vec{c}|^2 = 120\).

Check the truth of options A and B

With \( |\vec{c}|^2 = 120 \) we can now check the truth of options A and B. For option A, we calculate \((120/2) - 12\), and for option B, we calculate \((120/2) + 12\) to see if they equal 12 and 30, respectively.

Evaluate the cross product formula in option C

Evaluate the cross product formula given by option C, \(|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}| = 48 \sqrt{3}\). This involves assessing whether or not the magnitude of the cross product expressions matches the given number.

Verify option D using the dot product

To verify option D, we calculate the dot product \(\vec{a} \cdot \vec{b}\) using the magnitudes and the angle between them. Since \(\vec{a}\) and \(\vec{b}\) form a side and an extension of a side of the triangle, their dot product is negative. Use the relation \(\vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cos(180^\circ - \theta)\), where \(\theta\) is the angle opposite \(\vec{c}\).

Key concepts

Law of Cosines

The Law of Cosines is an extension of the Pythagorean theorem applicable to any triangle, not just right-angled ones. It relates the lengths of the sides of a triangle to the cosine of one of its angles.

In the context of vector algebra, especially for an exam like the JEE Advanced, the Law of Cosines can be expressed in terms of vectors and their dot product. For instance, If \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are vectors representing the sides of triangle PQR such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), the Law of Cosines can be reformulated as \(\vec{b} \cdot \vec{c} = |\vec{b}|^2 + |\vec{c}|^2 - |\vec{a}|^2\). This formula allows us to calculate the unknown magnitude or angle between vectors, thus providing a critical connection between geometric figures and vector quantities.

For students aiming to tackle problems effectively in JEE Advanced, understanding the application of this law to vectors is invaluable, as it often provides a direct route to solutions involving vector magnitudes and angles.

Dot Product

In vector algebra, the dot product or scalar product is a way to multiply two vectors resulting in a scalar, not a vector. It is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation combines the magnitude of the vectors and the cosine of the angle between them.

Mathematically, the dot product between two vectors \(\vec{A}\) and \(\vec{B}\) is denoted as \(\vec{A} \cdot \vec{B}\) and is calculated as \(\vec{A} \cdot \vec{B} = |\vec{A}| \cdot |\vec{B}| \cos(\theta)\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

In JEE Advanced problems, the dot product can be used to find the angle between two vectors or to test if vectors are perpendicular, as the dot product is zero if the vectors are orthogonal. Moreover, as we saw in the previous solution step, it can be applied using the Law of Cosines in vector form to find unknown vector magnitudes.

Cross Product

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input vectors, making it a pivotal concept in vector algebra for students to master for exams like the JEE Advanced.

The cross product of vectors \(\vec{A}\) and \(\vec{B}\) is denoted by \(\vec{A} \times \vec{B}\) and can be computed using the determinant of a matrix formed by unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\), and the components of \(\vec{A}\) and \(\vec{B}\). The magnitude of the cross product \(|\vec{A} \times \vec{B}|\) equates to \(|\vec{A}| \cdot |\vec{B}| \sin(\theta)\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

In solutions to problems, the cross product can be utilized to determine if two vectors are parallel (as the cross product will be \(\vec{0}\) ) or calculate the area of parallelograms defined by the vectors. A deeper comprehension of this product assists in solving geometric problems, specifically those concerning the direction and orientation in space.

Vector Magnitude

Vector magnitude is a fundamental concept that refers to the 'length' or 'size' of the vector. In the context of a JEE Advanced physics or mathematics exercise, understanding how to compute the magnitude is essential, as it is frequently part of vector operations like dot and cross products.

The magnitude of a vector \(\vec{A}\) is denoted as \(|\vec{A}|\) and can be found using the square root of the sum of the squares of its components in Cartesian coordinates. If \(\vec{A}\) has components \(A_x\), \(A_y\), and \(A_z\), then \(|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}\).

For a vector representing a side of a triangle in a question, such as in the exercise provided, calculating the magnitude is often the first step before applying formulas for dot or cross products, or the Law of Cosines. A clear grasp of how to find vector magnitude forms the bedrock upon which other more complex vector problems can be solved efficiently.