1. Problem: Given vector $v = (-1, 2, 5)$, find all scalars $k$ such that $||kv|| = 4$. Step 1: Recall that $||kv|| = |k| imes ||v||$. Step 2: Calculate $||v|| = \sqrt{(-1)^2 + 2^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$. Step 3: We want $|k| \sqrt{30} = 4$ so $|k| = \frac{4}{\sqrt{30}}$. Step 4: Thus $k = \pm \frac{4}{\sqrt{30}}$. 2. Problem: The norm of $u = (-3, k, 2)$ is 5; find $k$. Step 1: $||u|| = \sqrt{(-3)^2 + k^2 + 2^2} = 5$. Step 2: Calculate $\sqrt{9 + k^2 + 4} = 5$. Step 3: Squaring both sides, $9 + k^2 + 4 = 25$ so $k^2 = 25 - 13 = 12$. Step 4: Thus $k = \pm \sqrt{12} = \pm 2\sqrt{3}$. 3. Problem: Find $x,y$ such that $v = (x,y,3)$ has norm 14 and is orthogonal to $w = (3,3,3)$. Step 1: Orthogonality implies $v \cdot w = 0$ so $3x + 3y + 9 = 0$ or $x + y = -3$. Step 2: Norm condition $||v|| = \sqrt{x^2 + y^2 + 9} = 14$. Step 3: Square both sides: $x^2 + y^2 + 9 = 196$, so $x^2 + y^2 = 187$. Step 4: Using $y = -3 - x$, substitute: $$x^2 + (-3 - x)^2 = 187$$ $$x^2 + (x^2 + 6x + 9) = 187$$ $$2x^2 + 6x + 9 = 187$$ $$2x^2 + 6x - 178 = 0$$ Step 5: Divide by 2: $$x^2 + 3x - 89 = 0$$ Step 6: Solve quadratic: $$x = \frac{-3 \pm \sqrt{9 + 356}}{2} = \frac{-3 \pm \sqrt{365}}{2}$$ Step 7: Then $y = -3 - x$. 4. Problem: Show line joining midpoints of two sides of triangle is parallel to third side and half its length. Step 1: Let triangle vertices be $A,B,C$; midpoints of $AB$ and $AC$ are $M=\frac{A+B}{2}$ and $N=\frac{A+C}{2}$. Step 2: Vector $MN = N - M = \frac{A+C}{2} - \frac{A+B}{2} = \frac{C-B}{2}$. Step 3: $C - B$ is vector representing side $BC$; thus $MN$ is parallel to $BC$ and has length half of $BC$. 5. Problem: Prove that quadrilateral $PQRS$ with vertices midpoints of sides of quadrilateral $ABCD$ is a parallelogram. Step 1: Let $P, Q, R, S$ be midpoints of $AB, BC, CD, DA$ respectively. Step 2: Use midpoint formula for each point. Step 3: Vector $PQ = Q - P = \frac{B + C}{2} - \frac{A + B}{2} = \frac{C - A}{2}$. Step 4: Vector $SR = R - S = \frac{C + D}{2} - \frac{D + A}{2} = \frac{C - A}{2}$. Step 5: Since $PQ = SR$, sides are equal and parallel. Step 6: Similarly prove $QR = SP$; therefore $PQRS$ is parallelogram. 6. Problem: Using vector show angle inscribed in a semicircle is right angle. Step 1: Let diameter endpoints be $A$, $B$ and point on semicircle be $C$. Step 2: Vectors: $u = A - C$, $v = B - C$. Step 3: Show $u \cdot v = 0$. Step 4: Since $AB$ is diameter and $C$ lies on circle, using circle properties, $AC^2 + BC^2 = AB^2$. Step 5: Dot product $u \cdot v = 0$ implies angle at $C$ is right angle. 7. Problem: Show if $u$ is orthogonal to $v$ and $w$, then $u$ is orthogonal to $v + w$. Step 1: Since $u$ orthogonal to $v$, $u \cdot v = 0$. Step 2: Since $u$ orthogonal to $w$, $u \cdot w = 0$. Step 3: Then $u \cdot (v + w) = u \cdot v + u \cdot w = 0 + 0 = 0$. 8. Problem: Point $P$ divides segment $AB$ in ratio 3:4; $A, B$ have position vectors $u,v$; find position vector of $P$. Step 1: Using section formula, $$P = \frac{4u + 3v}{3 + 4} = \frac{4u + 3v}{7}$$ 9. Problem: Given $u = (1,1,2), v = (4,k,-3), w = (3,2,-1)$ are coplanar; find (a) $\mathrm{Proj}_u^v$, (b) $\mathrm{Proj}_u^w$, (c) $\mathrm{Proj}_u^{2u+v}$. Step 1: Coplanarity implies scalar triple product is zero: $$u \cdot (v \times w) = 0$$. Step 2: Calculate $v \times w = (k \times -1 - (-3)\times 2, -3 \times 3 - 4 \times (-1), 4 \times 2 - k \times 3)$, compute carefully. Step 3: Use formula for projection of $a$ onto $b$: $$\mathrm{Proj}_b^a = \frac{a \cdot b}{||b||^2} b$$. Step 4: Use values to compute each requested projection. 10. Problem: Given $u = (2,-1)$ and $v = (-1,2)$, find $w$ such that $w \cdot u = 1$ and $w \cdot v = 0$. Step 1: Write $w = (w_1, w_2)$. Step 2: Equations: $$2w_1 - w_2 = 1$$ $$-w_1 + 2w_2 = 0$$ Step 3: Solve system to find $w_1, w_2$. 11. Problem: (i) Find cosine angle between $u=(4,1,6)$ and $v=(-3,0,2)$. Step 1: Compute dot product $u \cdot v = 4 \times -3 + 1 \times 0 + 6 \times 2 = -12 + 0 + 12 = 0$. Step 2: Compute norms $||u||$ and $||v||$. Step 3: Cosine angle $= \frac{u\cdot v}{||u||\times ||v||} = 0$ implies angle is $\pi/2$. (ii) Find unit vector orthogonal to $u = (-1,3,2)$ and $v = (0,1,1)$. Step 1: Find cross product $u \times v$. Step 2: Normalize vector to unit length. 12. Problem: Triangle vertices $A(-1,2,1), B(3,1,1), C(2,-2,5)$. (i) Find side lengths. Step 1: Compute vectors $AB$, $BC$, $CA$. Step 2: Calculate their norms. (ii) Find interior angles using dot product formula. (iii) Find area via half magnitude of cross product of two sides. 13. Problem: Angles between $u,v,w$ are $\pi /3$, $||u||=4$, $||v||=1$, $||w||=8$; find $||u+v+w||$. Step 1: Use formula: $$||u+v+w||^2 = ||u||^2 + ||v||^2 + ||w||^2 + 2(u\cdot v + v\cdot w + w\cdot u)$$ Step 2: Use cosine and magnitudes to find dot products. Step 3: Substitute and compute norm. 14. Problem: $w+u+v=0$, $||u||=3, ||v||=5, ||w||=7$, find angle between $u$ and $v$. Step 1: Square both sides: $$||w + u + v||^2 = 0$$ $$||w||^2 + ||u||^2 + ||v||^2 + 2(w\cdot u + w\cdot v + u\cdot v) = 0$$ Step 2: Since $w = - (u+v)$, use substitution to find $u \cdot v$. Step 3: Use $u \cdot v = ||u|| ||v|| \cos \theta$ to find $\theta$. 15. Problem: Show $d(u,v) = ||u-v||$ satisfies: (i) Symmetry: $d(u,v) = d(v,u)$ (ii) Triangle inequality: $d(u,v) \le d(u,w) + d(w,v)$ Step 1: Use properties of norm (Euclidean distance). Step 2: Symmetry is true since $||u-v|| = ||v-u||$. Step 3: Triangle inequality follows from norm properties. Final answer available in steps above for each question.