🤖 AI Generated Solution:

**Correct Option if MCQ:** B **Reasoning:** * Calculate the dot product: \(\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1) = 3 - 2 - 1 = 0\). * Since the dot product is zero, the vectors are perpendicular. * \(\vec{a} \perp \vec{b}\)

✅ Manual Solution:

\((\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}})\vec{b}\)

✅ Manual Solution:

\(\frac{2\pi}{3}\)

✅ Manual Solution:

The position vector of the midpoint \(D\) is the average of the position vectors of \(B\) and \(C\) relative to \(A\):\[\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\] \[\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\] The length of the median is the magnitude of the vector \(\vec{AD}\), denoted as \(|\vec{AD}|\):\[|\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(0\right)^2 + \left(\frac{5}{2}\right)^2}\] \[|\vec{AD}| = \frac{\sqrt{34}}{2}\]

🤖 AI Generated Solution:

**Correct Option if MCQ:** N/A **Reasoning:** * \(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\) * \(\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12\) * \(|\vec{a}| = \sqrt{3^2+1^2+2^2} = \sqrt{14}\), \(|\vec{b}| = \sqrt{2^2+(-2)^2+4^2} = \sqrt{24}\). \(\theta = \arccos\left(\frac{12}{\sqrt{14}\sqrt{24}}\right) = \arccos\left(\frac{12}{\sqrt{336}}\right)\)

✅ Manual Solution:

Let \(\alpha\), \(\beta\), and \(\gamma\) be the angles the vector \(\vec{a}\) makes with the positive \(x\), \(y\), and \(z\) axes, respectively. The direction cosines are \(l = \cos\alpha\), \(m = \cos\beta\), and \(n = \cos\gamma\).Since the vector makes equal angles with all three axes, we have \(\alpha = \beta = \gamma\).Therefore, the direction cosines are equal: \(l = m = n\).The fundamental identity for direction cosines is:\[l^2 + m^2 + n^2 = 1\]Substituting \(l = m = n\):\[l^2 + l^2 + l^2 = 1\]\[3l^2 = 1\]\[l^2 = \frac{1}{3}\]\[l = \pm\frac{1}{\sqrt{3}}\]So, the direction cosines are \(l = m = n = \pm\frac{1}{\sqrt{3}}\).

A vector \(\vec{a}\) can be expressed in terms of its magnitude \(|\vec{a}|\) and its direction cosines \((l, m, n)\) as:\[\vec{a} = |\vec{a}| (l\hat{i} + m\hat{j} + n\hat{k})\]Given the magnitude of the vector is \(|\vec{a}| = 5\sqrt{3}\) units.

Case 1: Using the positive sign for the direction cosines, \(l = m = n = \frac{1}{\sqrt{3}}\).\[\vec{a}_1 = 5\sqrt{3} \left(\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\right)\]\[\vec{a}_1 = 5\sqrt{3} \cdot \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})\]\[\vec{a}_1 = 5 (\hat{i} + \hat{j} + \hat{k}) = 5\hat{i} + 5\hat{j} + 5\hat{k}\]

Case 2: Using the negative sign for the direction cosines, \(l = m = n = -\frac{1}{\sqrt{3}}\).\[\vec{a}_2 = 5\sqrt{3} \left(-\frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}\right)\]\[\vec{a}_2 = 5\sqrt{3} \cdot \left(-\frac{1}{\sqrt{3}}\right) (\hat{i} + \hat{j} + \hat{k})\]\[\vec{a}_2 = -5 (\hat{i} + \hat{j} + \hat{k}) = -5\hat{i} - 5\hat{j} - 5\hat{k}\]

Hence the vector \(\vec{a}\) that makes equal angles with all three axes and has a magnitude of \(5\sqrt{3}\) units is:\[\vec{a} = \mathbf{5\hat{i} + 5\hat{j} + 5\hat{k}}\]or\[\vec{a} = \mathbf{-5\hat{i} - 5\hat{j} - 5\hat{k}}\]

✅ Manual Solution:

If two vectors are collinear, one is a scalar multiple of the other.Let \(\vec{\alpha }=k\vec{\beta }\).

Substituting in \((x-2)\vec{a}+\vec{b}=k[(3+2x)\vec{a}-2\vec{b}]\)

\((x-2)\vec{a}+\vec{b}=k(3+2x)\vec{a}-2k\vec{b}\)

Since \(\vec{a}\) and \(\vec{b}\) are non-collinear, equating the coefficients of a and b both sides we get

\(x-2=k(3+2x)\) (Eq. 1) and

\(1=-2k\) (Eq. 2)

Solving for \(k\) and \(x\) we get \(k=-\frac{1}{2}\).

Substitute \(k\) into Eq. 1 we get

\(x-2=-\frac{1}{2}(3+2x)\)

\(2(x-2)=-(3+2x)\)

\(2x-4=-3-2x\)

\(4x=1\)

\(x=\frac{1}{4}\)

🤖 AI Generated Solution:

Let the coordinates of the endpoints of the wire be $A(x_1, y_1, z_1) = A(4,1,-2)$ and $B(x_2, y_2, z_2) = B(6,2,-3)$. To trisect the wire AB, we need two points, $P_1$ and $P_2$. $P_1$ divides AB in the ratio $1:2$, and $P_2$ divides AB in the ratio $2:1$. Using the section formula for a point $P(x,y,z)$ dividing a line segment in ratio $m:n$: $$P = \left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}, \frac{nz_1 + mz_2}{m+n} \right)$$ For the first point $P_1$ (ratio $m:n = 1:2$): $$P_1 = \left( \frac{2(4) + 1(6)}{1+2}, \frac{2(1) + 1(2)}{1+2}, \frac{2(-2) + 1(-3)}{1+2} \right)$$ $$P_1 = \left( \frac{8+6}{3}, \frac{2+2}{3}, \frac{-4-3}{3} \right) = \left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right)$$ For the second point $P_2$ (ratio $m:n = 2:1$): $$P_2 = \left( \frac{1(4) + 2(6)}{2+1}, \frac{1(1) + 2(2)}{2+1}, \frac{1(-2) + 2(-3)}{2+1} \right)$$ $$P_2 = \left( \frac{4+12}{3}, \frac{1+4}{3}, \frac{-2-6}{3} \right) = \left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right)$$ The coordinates of the points where the lanterns are hung are $\left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right)$ and $\left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right)$.

✅ Manual Solution:

Let \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) be the position vectors of points P, Q, and R, respectively. The vector from Q to P is found by subtracting the position vector of the tail (Q) from the position vector of the head (P):\[\vec{QP} = \vec{p} - \vec{q} = \vec{\alpha} - \vec{\beta}\] Given that \[\vec{QR} = \frac{3}{2} \vec{QP}\] Substitute the expression for \(\vec{QP}\):\[\vec{QR} = \frac{3}{2} (\vec{\alpha} - \vec{\beta})\] \[\vec{r} = \vec{\beta} + \frac{3}{2} (\vec{\alpha} - \vec{\beta})\] \[\vec{r} = \frac{3}{2} \vec{\alpha} + \left(1 - \frac{3}{2}\right) \vec{\beta}\] The position vector of the point R is:$\(\vec{r} = \frac{3}{2} \vec{\alpha} - \frac{1}{2} \vec{\beta}\)$

🤖 AI Generated Solution:

Hello Class 12 students! Let's tackle this interesting vector problem. We need to find a vector with a specific magnitude that is perpendicular to two given vectors. **Understanding the Problem** We are given two vectors: \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) \(\vec{b} = \hat{i} + \hat{j} - \hat{k}\) We need to find a vector, let's call it \(\vec{v}\), such that: 1. The magnitude of \(\vec{v}\) is 4 units (i.e., \(|\vec{v}| = 4\)). 2. \(\vec{v}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). **Key Concept: Cross Product** A vector that is perpendicular to two other vectors is given by their cross product. The cross product of \(\vec{a}\) and \(\vec{b}\), denoted by \(\vec{a} \times \vec{b}\), results in a vector that is orthogonal (perpendicular) to both \(\vec{a}\) and \(\vec{b}\). **Step 1: Calculate the Cross Product \(\vec{a} \times \vec{b}\)** The cross product of two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) can be calculated using the determinant of a matrix: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$ For our vectors, \(a_1=2, a_2=-1, a_3=1\) and \(b_1=1, b_2=1, b_3=-1\). $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} $$ Expanding the determinant: $$ \vec{a} \times \vec{b} = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(1) - (-1)(1)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(1 - 1) - \hat{j}(-2 - 1) + \hat{k}(2 - (-1)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(0) - \hat{j}(-3) + \hat{k}(3) $$ $$ \vec{a} \times \vec{b} = 0\hat{i} + 3\hat{j} + 3\hat{k} $$ So, a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is \(3\hat{j} + 3\hat{k}\). **Step 2: Find the Magnitude of the Perpendicular Vector** Let \(\vec{p} = \vec{a} \times \vec{b} = 3\hat{j} + 3\hat{k}\). The magnitude of \(\vec{p}\) is given by: \(|\vec{p}| = \sqrt{p_1^2 + p_2^2 + p_3^2}\) $$ |\vec{p}| = \sqrt{(0)^2 + (3)^2 + (3)^2} = \sqrt{0 + 9 + 9} = \sqrt{18} $$ We can simplify \(\sqrt{18}\) as \(\sqrt{9 \times 2} = 3\sqrt{2}\). So, \(|\vec{p}| = 3\sqrt{2}\) units. **Step 3: Find a Unit Vector in the Direction of \(\vec{p}\)** To get a vector of a specific magnitude, we first find a unit vector in the direction of our perpendicular vector \(\vec{p}\). A unit vector is found by dividing the vector by its magnitude. Let \(\hat{p}\) be the unit vector: $$ \hat{p} = \frac{\vec{p}}{|\vec{p}|} = \frac{3\hat{j} + 3\hat{k}}{3\sqrt{2}} $$ $$ \hat{p} = \frac{3}{3\sqrt{2}}\hat{j} + \frac{3}{3\sqrt{2}}\hat{k} $$ $$ \hat{p} = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} $$ **Step 4: Find the Vector of Magnitude 4 Units** We need a vector \(\vec{v}\) with a magnitude of 4 units in the direction of \(\hat{p}\). We can obtain this by multiplying the unit vector \(\hat{p}\) by the desired magnitude (4). $$ \vec{v} = 4 \times \hat{p} $$ $$ \vec{v} = 4 \left( \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \right) $$ $$ \vec{v} = \frac{4}{\sqrt{2}}\hat{j} + \frac{4}{\sqrt{2}}\hat{k} $$ To rationalize the denominators: $$ \vec{v} = \frac{4\sqrt{2}}{2}\hat{j} + \frac{4\sqrt{2}}{2}\hat{k} $$ $$ \vec{v} = 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k} $$ **Important Note:** The cross product is not unique in direction. If \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\), then \(-\vec{a} \times \vec{b}\) is also perpendicular. So, we also have another possible vector: $$ \vec{v}' = -4 \times \hat{p} = -2\sqrt{2}\hat{j} - 2\sqrt{2}\hat{k} $$ Both vectors are valid answers. Let's choose \(\vec{v} = 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}\) for verification. **Step 5: Verify the Answer** We need to verify two things for our vector \(\vec{v} = 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}\): **Verification 1: Magnitude is 4 units.** $$ |\vec{v}| = \sqrt{(0)^2 + (2\sqrt{2})^2 + (2\sqrt{2})^2} $$ $$ |\vec{v}| = \sqrt{0 + (4 \times 2) + (4 \times 2)} $$ $$ |\vec{v}| = \sqrt{0 + 8 + 8} = \sqrt{16} = 4 $$ The magnitude is indeed 4 units. **Verification 2: Perpendicular to \(\vec{a}\) and \(\vec{b}\).** A vector is perpendicular to another if their dot product is zero. **Check perpendicularity with \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\):** $$ \vec{v} \cdot \vec{a} = (0\hat{i} + 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k}) $$ $$ \vec{v} \cdot \vec{a} = (0)(2) + (2\sqrt{2})(-1) + (2\sqrt{2})(1) $$ $$ \vec{v} \cdot \vec{a} = 0 - 2\sqrt{2} + 2\sqrt{2} = 0 $$ So, \(\vec{v}\) is perpendicular to \(\vec{a}\). **Check perpendicularity with \(\vec{b} = \hat{i} + \hat{j} - \hat{k}\):** $$ \vec{v} \cdot \vec{b} = (0\hat{i} + 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) $$ $$ \vec{v} \cdot \vec{b} = (0)(1) + (2\sqrt{2})(1) + (2\sqrt{2})(-1) $$ $$ \vec{v} \cdot \vec{b} = 0 + 2\sqrt{2} - 2\sqrt{2} = 0 $$ So, \(\vec{v}\) is perpendicular to \(\vec{b}\). Both conditions are satisfied. **Final Answer:** A vector of magnitude 4 units perpendicular to each of the vectors \(2\hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}-\hat{k}\) is \(\boxed{2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}}\). The other possible vector is \(-2\sqrt{2}\hat{j} - 2\sqrt{2}\hat{k}\).

🤖 AI Generated Solution:

Here's a step-by-step solution to find the angles of $\Delta$ ABC: **Understanding the Problem:** We are given the position vectors of the vertices of a triangle ABC. We need to find the angles at each vertex (A, B, and C). To do this, we will first find the vectors representing the sides of the triangle and then use the dot product formula to find the angles. **Step 1: Find the vectors representing the sides of the triangle.** The vector from point A to point B ($\vec{AB}$) is given by the position vector of B minus the position vector of A. $\vec{AB} = \vec{OB} - \vec{OA}$ $\vec{AB} = (\hat{i}-3\hat{j}-5\hat{k}) - (2\hat{i}-\hat{j}+\hat{k})$ $\vec{AB} = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k}$ $\vec{AB} = -\hat{i} - 2\hat{j} - 6\hat{k}$ The vector from point B to point C ($\vec{BC}$) is given by the position vector of C minus the position vector of B. $\vec{BC} = \vec{OC} - \vec{OB}$ $\vec{BC} = (3\hat{i}-4\hat{j}-4\hat{k}) - (\hat{i}-3\hat{j}-5\hat{k})$ $\vec{BC} = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k}$ $\vec{BC} = 2\hat{i} - \hat{j} + \hat{k}$ The vector from point C to point A ($\vec{CA}$) is given by the position vector of A minus the position vector of C. $\vec{CA} = \vec{OA} - \vec{OC}$ $\vec{CA} = (2\hat{i}-\hat{j}+\hat{k}) - (3\hat{i}-4\hat{j}-4\hat{k})$ $\vec{CA} = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k}$ $\vec{CA} = -\hat{i} + 3\hat{j} + 5\hat{k}$ **Step 2: Find the magnitudes of the side vectors.** The magnitude of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$. $|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$ $|\vec{BC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ $|\vec{CA}| = \sqrt{(-1)^2 + (3)^2 + (5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$ **Step 3: Calculate the angles using the dot product.** The dot product of two vectors $\vec{u}$ and $\vec{v}$ is given by $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$, where $\theta$ is the angle between them. Therefore, $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$. **Angle A (Angle BAC):** This angle is formed by vectors $\vec{AB}$ and $\vec{AC}$. We need to use $\vec{AB}$ and $-\vec{CA}$ (which is $\vec{AC}$). $\vec{AC} = -\vec{CA} = -(-\hat{i} + 3\hat{j} + 5\hat{k}) = \hat{i} - 3\hat{j} - 5\hat{k}$ Alternatively, $\vec{AC} = \vec{OC} - \vec{OA} = (3\hat{i}-4\hat{j}-4\hat{k}) - (2\hat{i}-\hat{j}+\hat{k}) = \hat{i} - 3\hat{j} - 5\hat{k}$. $\vec{AB} \cdot \vec{AC} = (-\hat{i} - 2\hat{j} - 6\hat{k}) \cdot (\hat{i} - 3\hat{j} - 5\hat{k})$ $\vec{AB} \cdot \vec{AC} = (-1)(1) + (-2)(-3) + (-6)(-5)$ $\vec{AB} \cdot \vec{AC} = -1 + 6 + 30 = 35$ $|\vec{AB}| = \sqrt{41}$ $|\vec{AC}| = \sqrt{(1)^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$ $\cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}||\vec{AC}|} = \frac{35}{\sqrt{41}\sqrt{35}} = \frac{\sqrt{35}}{\sqrt{41}} = \sqrt{\frac{35}{41}}$ $A = \cos^{-1}\left(\sqrt{\frac{35}{41}}\right)$ **Angle B (Angle ABC):** This angle is formed by vectors $\vec{BA}$ and $\vec{BC}$. $\vec{BA} = -\vec{AB} = -(-\hat{i} - 2\hat{j} - 6\hat{k}) = \hat{i} + 2\hat{j} + 6\hat{k}$ $\vec{BA} \cdot \vec{BC} = (\hat{i} + 2\hat{j} + 6\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})$ $\vec{BA} \cdot \vec{BC} = (1)(2) + (2)(-1) + (6)(1)$ $\vec{BA} \cdot \vec{BC} = 2 - 2 + 6 = 6$ $|\vec{BA}| = |\vec{AB}| = \sqrt{41}$ $|\vec{BC}| = \sqrt{6}$ $\cos B = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}||\vec{BC}|} = \frac{6}{\sqrt{41}\sqrt{6}} = \frac{6}{\sqrt{246}}$ $B = \cos^{-1}\left(\frac{6}{\sqrt{246}}\right)$ **Angle C (Angle BCA):** This angle is formed by vectors $\vec{CB}$ and $\vec{CA}$. $\vec{CB} = -\vec{BC} = -(2\hat{i} - \hat{j} + \hat{k}) = -2\hat{i} + \hat{j} - \hat{k}$ $\vec{CB} \cdot \vec{CA} = (-2\hat{i} + \hat{j} - \hat{k}) \cdot (-\hat{i} + 3\hat{j} + 5\hat{k})$ $\vec{CB} \cdot \vec{CA} = (-2)(-1) + (1)(3) + (-1)(5)$ $\vec{CB} \cdot \vec{CA} = 2 + 3 - 5 = 0$ Since the dot product is 0, the angle between $\vec{CB}$ and $\vec{CA}$ is 90 degrees. $\cos C = \frac{\vec{CB} \cdot \vec{CA}}{|\vec{CB}||\vec{CA}|} = \frac{0}{\sqrt{6}\sqrt{35}} = 0$ $C = \cos^{-1}(0) = 90^\circ$ **Step 4: Verification (Optional but Recommended)** The sum of the angles in a triangle should be $180^\circ$. We have found $C = 90^\circ$. This means that $\Delta$ ABC is a right-angled triangle. Let's calculate the approximate values for angles A and B to check if they add up to $90^\circ$. $\cos A = \sqrt{\frac{35}{41}} \approx \sqrt{0.8536} \approx 0.9239$ $A \approx \cos^{-1}(0.9239) \approx 22.49^\circ$ $\cos B = \frac{6}{\sqrt{246}} \approx \frac{6}{15.68} \approx 0.3826$ $B \approx \cos^{-1}(0.3826) \approx 67.51^\circ$ $A + B + C \approx 22.49^\circ + 67.51^\circ + 90^\circ = 180^\circ$. The values are consistent. **Final Answer:** The angles of $\Delta$ ABC are: Angle A (angle BAC) = $\cos^{-1}\left(\sqrt{\frac{35}{41}}\right)$ Angle B (angle ABC) = $\cos^{-1}\left(\frac{6}{\sqrt{246}}\right)$ Angle C (angle BCA) = $90^\circ$

🤖 AI Generated Solution:

Hello Class 12! Today we're going to tackle a question about the area of a parallelogram using vectors. Let's break it down step-by-step. **Part 1: Showing the Formula for the Area of a Parallelogram** **1. Understanding the Geometry of a Parallelogram and its Diagonals** Let the parallelogram be denoted by ABCD. Let the adjacent sides be represented by vectors $\vec{u}$ and $\vec{v}$. The diagonals of the parallelogram are $\vec{a}$ and $\vec{b}$. From the parallelogram law of vector addition, we can express the diagonals in terms of the adjacent sides: $$ \vec{a} = \vec{u} + \vec{v} \quad (*)$$ $$ \vec{b} = \vec{u} - \vec{v} \quad (**)$$ **2. Relating the Diagonals to the Adjacent Sides** Our goal is to express the area of the parallelogram in terms of $\vec{a}$ and $\vec{b}$. We know that the area of a parallelogram formed by adjacent sides $\vec{u}$ and $\vec{v}$ is given by the magnitude of their cross product: Area $= |\vec{u} \times \vec{v}|$ Let's try to express $\vec{u}$ and $\vec{v}$ in terms of $\vec{a}$ and $\vec{b}$ by manipulating equations (*) and (**). Adding (*) and (**): $$ \vec{a} + \vec{b} = (\vec{u} + \vec{v}) + (\vec{u} - \vec{v}) $$ $$ \vec{a} + \vec{b} = 2\vec{u} $$ $$ \vec{u} = \frac{1}{2}(\vec{a} + \vec{b}) $$ Subtracting (**) from (*): $$ \vec{a} - \vec{b} = (\vec{u} + \vec{v}) - (\vec{u} - \vec{v}) $$ $$ \vec{a} - \vec{b} = 2\vec{v} $$ $$ \vec{v} = \frac{1}{2}(\vec{a} - \vec{b}) $$ **3. Calculating the Cross Product of Adjacent Sides** Now, let's find the cross product $\vec{u} \times \vec{v}$ using the expressions we found for $\vec{u}$ and $\vec{v}$: $$ \vec{u} \times \vec{v} = \left(\frac{1}{2}(\vec{a} + \vec{b})\right) \times \left(\frac{1}{2}(\vec{a} - \vec{b})\right) $$ Using the distributive property of the cross product: $$ \vec{u} \times \vec{v} = \frac{1}{4} [(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})] $$ $$ \vec{u} \times \vec{v} = \frac{1}{4} [\vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b}] $$ **4. Using Properties of the Cross Product** We know that the cross product of a vector with itself is the zero vector: $$ \vec{a} \times \vec{a} = \vec{0} $$ $$ \vec{b} \times \vec{b} = \vec{0} $$ Also, the cross product is anti-commutative, meaning $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. Substituting these properties back into the equation: $$ \vec{u} \times \vec{v} = \frac{1}{4} [\vec{0} - \vec{a} \times \vec{b} + (-\vec{a} \times \vec{b}) - \vec{0}] $$ $$ \vec{u} \times \vec{v} = \frac{1}{4} [-2(\vec{a} \times \vec{b})] $$ $$ \vec{u} \times \vec{v} = -\frac{1}{2}(\vec{a} \times \vec{b}) $$ **5. Finding the Area of the Parallelogram** The area of the parallelogram is the magnitude of $\vec{u} \times \vec{v}$: Area $= |\vec{u} \times \vec{v}|$ Area $= \left|-\frac{1}{2}(\vec{a} \times \vec{b})\right|$ Using the property that $|c\vec{v}| = |c||\vec{v}|$ for a scalar $c$: Area $= \frac{1}{2} |\vec{a} \times \vec{b}|$ Thus, we have shown that the area of a parallelogram whose diagonals are represented by $\vec{a}$ and $\vec{b}$ is given by $\frac{1}{2}|\vec{a}\times\vec{b}|$. --- **Part 2: Finding the Area of a Parallelogram with Given Diagonals** Given the diagonals: $$ \vec{a} = 2\hat{i}-\hat{j}+\hat{k} $$ $$ \vec{b} = \hat{i}+3\hat{j}-\hat{k} $$ **1. Calculate the Cross Product $\vec{a} \times \vec{b}$** We will use the determinant method to calculate the cross product: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} $$ $$ \vec{a} \times \vec{b} = \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 - (-1)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(-2) - \hat{j}(-3) + \hat{k}(7) $$ $$ \vec{a} \times \vec{b} = -2\hat{i} + 3\hat{j} + 7\hat{k} $$ **2. Calculate the Magnitude of the Cross Product** Now, we find the magnitude of the resulting vector: $$ |\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2} $$ $$ |\vec{a} \times \vec{b}| = \sqrt{4 + 9 + 49} $$ $$ |\vec{a} \times \vec{b}| = \sqrt{62} $$ **3. Calculate the Area of the Parallelogram** Using the formula derived in Part 1: Area $= \frac{1}{2} |\vec{a} \times \vec{b}|$ Area $= \frac{1}{2} \sqrt{62}$ So, the area of the parallelogram whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$ is $\frac{\sqrt{62}}{2}$ square units. I hope this explanation was clear and helpful! Let me know if you have any more questions.