Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I I. The points \(O, A, B, C, D\) are such that \(O A=a, O B=b, O C=2 a+3 b\) and \(O D\) \(=a-2 b\). If \(|a|=3|b|\), then the angle between \(B D\) and \(A C\) is II. \(a, b, c\) are three unit vectors such that \(a \times(b \times c)=\frac{1}{2}(b+c) .\) If the vectors \(b\) and \(c\) are non-parallel, then the angle between \(a\) and \(b\) is III. Let the vectors \(a, b, c\) and \(d\) be such that \((a \times b) \times(c \times d)=0\). Let \(P_{1}\) and \(P_{2}\) be planes determined by the pairs of vectors \(a, b\) and \(c, d\) respectively, then the angle between \(P_{1}\) and \(P_{2}\) is IV. If \(a\) and \(b\) are two vectors such that \(a \cdot b<0\) and \(|a \cdot b|=|a \times b|\), then the angle between vectors \(a\) and \(b\) is Column-I (A) 0 (B) \(\frac{2 \pi}{3}\) (C) \(\frac{3 \pi}{4}\) (D) \(\frac{\pi}{2}\)

Step-by-Step Solution

Answer

(I) \(\frac{\pi}{2}\), (II) \(\frac{2 \pi}{3}\), (III) 0, (IV) \(\frac{3 \pi}{4}\)

Step 1: Analyze the information for I

The vectors are given as \(OA = a\), \(OB = b\), \(OC = 2a + 3b\), and \(OD = a - 2b\). We need the angle between \(BD\) and \(AC\). First, find the vectors: \(BD = OD - OB = (a - 2b) - b = a - 3b\) and \(AC = OC - OA = (2a + 3b) - a = a + 3b\). Then, calculate the dot product: \((a - 3b) \cdot (a + 3b) = |a|^2 + (a \cdot 3b) - (3b \cdot a) - 9|b|^2\). If \(|a|=3|b|\), substitute \(|a|^2 = 9|b|^2\). Using \(a \cdot b = -|a||b| = -3|b|^2\), we can simplify the dot product to \(0\). Therefore, the angle is \(\frac{\pi}{2}\) since the dot product is zero.

Step 2: Solve for II

Using the vector identity \(a \times (b \times c) = (a \cdot c)b - (a \cdot b)c\), and given \(a \times(b \times c)=\frac{1}{2}(b+c)\), equate components: \((a \cdot c)b - (a \cdot b)c = \frac{1}{2}(b+c)\). By comparing components, \(a \cdot c = \frac{1}{2}\) and \(a \cdot b = -\frac{1}{2}\). The cosine of the angle between them is \(-\cos \theta = \frac{1}{2}\), hence \(\theta = \frac{2\pi}{3}\). Thus, the angle between \(a\) and \(b\) is \(\frac{2 \pi}{3}\).

Step 3: Examine condition for III

Given \((a \times b) \times(c \times d)=0\), this implies the planes determined by \((a, b)\) and \((c, d)\) share a common direction or are parallel. Thus, the angle between \(P_1\) and \(P_2\) is \(0\).

Step 4: Evaluate IV for angle conditions

We know \(a \cdot b < 0\) and \(|a \cdot b| = |a \times b|\). Comparing magnitudes, we obtain \((a \cdot b)^2 = |a|^2|b|^2 \sin^2 \theta\). Simplifying, \(|\cos \theta| = |\sin \theta|\), giving us \(\theta = \frac{3\pi}{4}\). Therefore, the angle is \(\frac{3 \pi}{4}\).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It is a way to multiply two vectors, resulting in a scalar. Attaching a numerical value to the interaction of directions and magnitudes of the vectors, the dot product is calculated as follows:

• The formula for the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\), where \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are the lengths or magnitudes of the vectors and \(\theta\) is the angle between them.
• The dot product is also equal to the sum of the products of their respective components if the vectors are in component form. For example, if \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), then \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\).

It gives us a measure of how much one vector extends in the direction of another. If the dot product is zero, it signifies that the vectors are perpendicular. In the given problem, the vectors \(\mathbf{BD}\) and \(\mathbf{AC}\) showed a dot product of zero, confirming their orthogonality. Understanding this concept is key to solving problems with vectors in geometry.

Cross Product

The cross product is another method of multiplying two vectors in three-dimensional space, resulting in a new vector that is perpendicular to both original vectors. The cross product is useful when you need to find a vector orthogonal to a plane formed by the two vectors.

• The formula for the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\). This results in a vector whose direction is determined by the right-hand rule and whose magnitude is given by \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
• In component form, if \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), calculating each component independently provides clarity.

The cross product is particularly relevant when dealing with torque, rotational dynamics, and determining the area of parallelograms in physics and engineering. In the problem, understanding the cross product was crucial when assessing the conditions of certain plane and vector arrangements.

Angle Between Vectors

Understanding the angle between vectors is pivotal in vector algebra, giving insight into their orientation relative to one another. The angle is crucial for determining various properties and interactions between vectors.

• The angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be found using the dot product formula: \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\). Solving for \(\theta\) involves using the inverse cosine function, i.e., \(\theta = \cos^{-1}(\cos \theta)\).
• In the problem, when the dot product of vectors was zero, it indicated that the vectors were perpendicular, hence the angle was \(\frac{\pi}{2}\). Conversely, when the dot product between two vectors was negative, it suggested an obtuse angle, such as \(\frac{3\pi}{4}\).
• If information on the cross product is involved, we can also look at \(\sin \theta\), using the relationship \(\sin \theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| |\mathbf{b}|}\).

By understanding both the dot product and cross product in relation to the angle \(\theta\), one can derive critical insights into vector dynamics and interactions in a graphical and practical sense.