Vardaan Learning Institute

Proof: LHS = $(\vec{a}+\vec{b}) \cdot ((\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})) = (\vec{a}+\vec{b}) \cdot (\vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{c} + \vec{c}\times\vec{a})$. Since $\vec{c}\times\vec{c}=0$, this becomes $(\vec{a}+\vec{b}) \cdot (\vec{b}\times\vec{c} - \vec{a}\times\vec{b} + \vec{c}\times\vec{a})$. Distributing the dot product: $\vec{a}\cdot(\vec{b}\times\vec{c}) - \vec{a}\cdot(\vec{a}\times\vec{b}) + \vec{a}\cdot(\vec{c}\times\vec{a}) + \vec{b}\cdot(\vec{b}\times\vec{c}) - \vec{b}\cdot(\vec{a}\times\vec{b}) + \vec{b}\cdot(\vec{c}\times\vec{a})$. Any STP with repeated vectors is 0. So, $[\vec{a}\vec{b}\vec{c}] - 0 + 0 + 0 - 0 + [\vec{b}\vec{c}\vec{a}] = [\vec{a}\vec{b}\vec{c}] + [\vec{a}\vec{b}\vec{c}] = 2[\vec{a}\vec{b}\vec{c}]$.

Proof: Using BAC-CAB rule: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$. Similarly, $\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b}\cdot\vec{a})\vec{c} - (\vec{b}\cdot\vec{c})\vec{a}$ and $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c}\cdot\vec{b})\vec{a} - (\vec{c}\cdot\vec{a})\vec{b}$. Summing these three gives: $( (\vec{a}\cdot\vec{c}) - (\vec{c}\cdot\vec{a}) )\vec{b} + ( (\vec{b}\cdot\vec{a}) - (\vec{a}\cdot\vec{b}) )\vec{c} + ( (\vec{c}\cdot\vec{b}) - (\vec{b}\cdot\vec{c}) )\vec{a} = \vec{0}$.

Ans: Angle with $\vec{b}$ is $\frac{3\pi}{4}$, Angle with $\vec{c}$ is $\frac{\pi}{4}$. Reason: $(\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = \frac{1}{\sqrt{2}}\vec{b} + \frac{1}{\sqrt{2}}\vec{c}$. Since $\vec{b}, \vec{c}$ are non-collinear, we compare coefficients: $\vec{a}\cdot\vec{c} = 1/\sqrt{2} \Rightarrow \cos\theta_c = 1/\sqrt{2} \Rightarrow \theta_c = \pi/4$. And $-(\vec{a}\cdot\vec{b}) = 1/\sqrt{2} \Rightarrow \cos\theta_b = -1/\sqrt{2} \Rightarrow \theta_b = 3\pi/4$.

Ans: $\vec{x} = \frac{\vec{b} + \vec{b}\times\vec{a} + (\vec{a}\cdot\vec{b})\vec{a}}{1+|\vec{a}|^2}$. Reason: Take dot product with $\vec{a}$: $\vec{x}\cdot\vec{a} + 0 = \vec{b}\cdot\vec{a}$. Next, take cross product with $\vec{a}$: $\vec{x}\times\vec{a} + (\vec{x}\times\vec{a})\times\vec{a} = \vec{b}\times\vec{a}$. Use BAC-CAB: $\vec{x}\times\vec{a} + (\vec{x}\cdot\vec{a})\vec{a} - (\vec{a}\cdot\vec{a})\vec{x} = \vec{b}\times\vec{a}$. From original eq, $\vec{x}\times\vec{a} = \vec{b} - \vec{x}$. Substitute this and $\vec{x}\cdot\vec{a} = \vec{b}\cdot\vec{a}$ to get: $(\vec{b} - \vec{x}) + (\vec{b}\cdot\vec{a})\vec{a} - |\vec{a}|^2\vec{x} = \vec{b}\times\vec{a}$. Solve for $\vec{x}$.

Proof: LHS = $(\vec{a}\times\vec{b}) \cdot ((\vec{b}\times\vec{c}) \times (\vec{c}\times\vec{a}))$. Let $\vec{u} = \vec{b}\times\vec{c}$. Then $\vec{u} \times (\vec{c}\times\vec{a}) = (\vec{u}\cdot\vec{a})\vec{c} - (\vec{u}\cdot\vec{c})\vec{a} = ([\vec{b}\vec{c}\vec{a}])\vec{c} - 0 = [\vec{a}\vec{b}\vec{c}]\vec{c}$. Substituting back: LHS = $(\vec{a}\times\vec{b}) \cdot ([\vec{a}\vec{b}\vec{c}]\vec{c}) = [\vec{a}\vec{b}\vec{c}] ((\vec{a}\times\vec{b})\cdot\vec{c}) = [\vec{a}\vec{b}\vec{c}] [\vec{a}\vec{b}\vec{c}] = [\vec{a}\vec{b}\vec{c}]^2$.

Proof: Since $\vec{c} \perp \vec{a}, \vec{b}$, $\vec{c}$ is parallel to $\vec{a}\times\vec{b}$. Since $|\vec{c}|=1$, $\vec{c} = \pm \frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$. Now, $[\vec{a}\vec{b}\vec{c}] = (\vec{a}\times\vec{b}) \cdot \vec{c} = \pm |\vec{a}\times\vec{b}|$. Squaring gives $[\vec{a}\vec{b}\vec{c}]^2 = |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2(\pi/6) = \frac{1}{4}|\vec{a}|^2|\vec{b}|^2$.

Ans: $\pm(2\hat{i} + 2\hat{k})$. Reason: Let $\vec{v} = x\vec{a} + y\vec{b} = (x+y)\hat{i} + (x-y)\hat{j} + (x+y)\hat{k}$. Orthogonality: $\vec{v}\cdot\vec{c} = 0 \Rightarrow (x+y)(1) + (x-y)(2) + (x+y)(-1) = 0 \Rightarrow x+y+2x-2y-x-y = 2x-2y=0 \Rightarrow x=y$. So $\vec{v} = x(2\hat{i}+2\hat{k})$. Magnitude $|\vec{v}|^2 = x^2(4+4) = 8x^2$. We want magnitude $\sqrt{8}$, so $8x^2 = 8 \Rightarrow x = \pm 1$.

Ans: $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$. Reason: Let $|\vec{a}|=|\vec{b}|=|\vec{c}|=m$. Let $\vec{s} = \vec{a}+\vec{b}+\vec{c}$. $|\vec{s}|^2 = 3m^2 + 0 \Rightarrow |\vec{s}| = m\sqrt{3}$. $\cos\theta = \frac{\vec{a}\cdot\vec{s}}{|\vec{a}||\vec{s}|} = \frac{\vec{a}\cdot\vec{a} + 0 + 0}{m(m\sqrt{3})} = \frac{m^2}{m^2\sqrt{3}} = \frac{1}{\sqrt{3}}$.

Proof: Let the origin be the orthocenter $H$, so $\vec{h} = \vec{0}$. The altitudes from A and B meet at H. So $\vec{a} \perp (\vec{c}-\vec{b}) \Rightarrow \vec{a}\cdot\vec{c} - \vec{a}\cdot\vec{b} = 0 \Rightarrow \vec{a}\cdot\vec{c} = \vec{a}\cdot\vec{b}$. Also $\vec{b} \perp (\vec{c}-\vec{a}) \Rightarrow \vec{b}\cdot\vec{c} = \vec{a}\cdot\vec{b}$. Therefore, $\vec{a}\cdot\vec{c} = \vec{b}\cdot\vec{c} \Rightarrow \vec{c}\cdot(\vec{a}-\vec{b}) = 0$. This means altitude from C passes through H, proving concurrency.

Proof: Let origin be center of circle. Let position vectors of diameter endpoints be $\vec{a}$ and $-\vec{a}$. Let $P$ on circle be $\vec{r}$. Then $|\vec{r}| = |\vec{a}|$. The vectors for the inscribed angle are $\vec{r}-\vec{a}$ and $\vec{r}-(-\vec{a}) = \vec{r}+\vec{a}$. Their dot product is $(\vec{r}-\vec{a})\cdot(\vec{r}+\vec{a}) = |\vec{r}|^2 - |\vec{a}|^2 = 0$. Hence they are perpendicular.

Proof: By definition, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. Taking absolute value: $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}||\cos\theta|$. Since the maximum value of $|\cos\theta|$ is 1, it follows that $|\vec{a} \cdot \vec{b}| \le |\vec{a}||\vec{b}|$. Equality holds when $\theta = 0$ or $\pi$ (vectors are collinear).

Proof: Expand: $(|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}) + (|\vec{b}|^2+|\vec{c}|^2-2\vec{b}\cdot\vec{c}) + (|\vec{c}|^2+|\vec{a}|^2-2\vec{c}\cdot\vec{a}) = 9$. Since they are unit vectors, $6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9 \Rightarrow 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -3$. Consider $|\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2(\sum\vec{a}\cdot\vec{b}) = 3 - 3 = 0$. Hence $\vec{a}+\vec{b}+\vec{c} = \vec{0}$.

Proof: Original volume $V = \frac{1}{6}|[\vec{a}\vec{b}\vec{c}]|$. New volume $V' = \frac{1}{6}|[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}]|$. From Q1, this STP equals $2[\vec{a}\vec{b}\vec{c}]$. Thus, $V' = \frac{1}{6}|2[\vec{a}\vec{b}\vec{c}]| = 2 \times (\frac{1}{6}|[\vec{a}\vec{b}\vec{c}]|) = 2V$.

Ans: $x = -\frac{1}{2}$. Reason: STP = $1(12+2x) - 0 + 1(2x^2-24) = 2x^2 + 2x - 12$. This is a quadratic in $x$ representing a parabola opening upwards. The minimum value occurs at $x = -\frac{b}{2a} = -\frac{2}{4} = -\frac{1}{2}$.

Ans: $\frac{\pi}{3}$. Reason: $\vec{c} \cdot \vec{d} = 0 \Rightarrow (\vec{a}+2\vec{b})\cdot(5\vec{a}-4\vec{b}) = 5|\vec{a}|^2 - 8|\vec{b}|^2 + 6\vec{a}\cdot\vec{b} = 0$. Since unit vectors, $5(1) - 8(1) + 6\cos\theta = 0 \Rightarrow -3 + 6\cos\theta = 0 \Rightarrow \cos\theta = 1/2$. Thus $\theta = \pi/3$.

Proof: The reciprocal vectors are defined as $\vec{p} = \frac{\vec{b}\times\vec{c}}{[\vec{a}\vec{b}\vec{c}]}$, $\vec{q} = \frac{\vec{c}\times\vec{a}}{[\vec{a}\vec{b}\vec{c}]}$, $\vec{r} = \frac{\vec{a}\times\vec{b}}{[\vec{a}\vec{b}\vec{c}]}$. Therefore, $\vec{a}\cdot\vec{p} = \frac{\vec{a}\cdot(\vec{b}\times\vec{c})}{[\vec{a}\vec{b}\vec{c}]} = \frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{a}\vec{b}\vec{c}]} = 1$. Similarly, $\vec{b}\cdot\vec{q} = 1$ and $\vec{c}\cdot\vec{r} = 1$. Sum is $1+1+1=3$.

Proof: Let $\vec{u} = \vec{a}\times\vec{b}$. LHS = $\vec{u} \cdot (\vec{c}\times\vec{d}) = [\vec{u} \, \vec{c} \, \vec{d}] = (\vec{u}\times\vec{c}) \cdot \vec{d} = ((\vec{a}\times\vec{b})\times\vec{c}) \cdot \vec{d}$. Apply BAC-CAB rule: $((\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}) \cdot \vec{d}$. Distributing dot product yields $(\vec{a}\cdot\vec{c})(\vec{b}\cdot\vec{d}) - (\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{d})$.

Ans: $|\vec{b}| = 1$. Reason: $\vec{c} = \vec{a}\times\vec{b} \Rightarrow \vec{c} \perp \vec{a}, \vec{b}$. $\vec{a} = \vec{b}\times\vec{c} \Rightarrow \vec{a} \perp \vec{b}, \vec{c}$. Hence $\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal. Magnitudes: $|\vec{c}| = |\vec{a}||\vec{b}|\sin(\pi/2) = |\vec{a}||\vec{b}|$. Also $|\vec{a}| = |\vec{b}||\vec{c}|$. Substitute $|\vec{c}|$ to get $|\vec{a}| = |\vec{b}|^2|\vec{a}|$. Since $\vec{a}\neq\vec{0}$, $|\vec{b}|^2 = 1 \Rightarrow |\vec{b}|=1$.

Proof: $|\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2\hat{a}\cdot\hat{b} = 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) = 4\sin^2(\theta/2)$. Taking square roots: $|\hat{a} - \hat{b}| = 2\sin(\theta/2) \Rightarrow \sin(\theta/2) = \frac{1}{2}|\hat{a}-\hat{b}|$.

Ans: $y = -2$. Reason: Vectors $\vec{AB} = (1, -1, 0)$, $\vec{AC} = (0, 1, -1)$, $\vec{AD} = (2, y-1, 1)$. For coplanarity, STP = $0$. $1(1+y-1) - (-1)(0 - (-2)) + 0 = 0 \Rightarrow y + 2 = 0 \Rightarrow y = -2$.

Proof: Expand the STP: $(\vec{a}-\vec{b}) \cdot ((\vec{b}-\vec{c}) \times (\vec{c}-\vec{a})) = (\vec{a}-\vec{b}) \cdot (\vec{b}\times\vec{c} - \vec{b}\times\vec{a} - \vec{c}\times\vec{c} + \vec{c}\times\vec{a}) = \vec{a}\cdot(\vec{b}\times\vec{c}) - \vec{b}\cdot(\vec{c}\times\vec{a}) = [\vec{a}\vec{b}\vec{c}] - [\vec{b}\vec{c}\vec{a}] = [\vec{a}\vec{b}\vec{c}] - [\vec{a}\vec{b}\vec{c}] = 0$.

Ans: $x = -1, y = -10, z = 8$. Reason: Dot products of any two must be 0. Let vectors be $\vec{u}, \vec{v}, \vec{w}$. $\vec{u}\cdot\vec{w} = 3x + 1 + 2 = 0 \Rightarrow 3x = -3 \Rightarrow x=-1$. Now $\vec{u} = -\hat{i}-\hat{j}+\hat{k}$. $\vec{u}\cdot\vec{v} = -2 - y - z = 0 \Rightarrow y+z = -2$. $\vec{v}\cdot\vec{w} = 6 - y - 2z = 0 \Rightarrow y+2z = 6$. Subtract equations: $z = 8 \Rightarrow y = -10$.

Proof: Let vertices be $\vec{a}, \vec{b}, \vec{c}$. Midpoint of BC is $\frac{\vec{b}+\vec{c}}{2}$. Point dividing median from A in $2:1$ is $\frac{2(\frac{\vec{b}+\vec{c}}{2}) + 1(\vec{a})}{2+1} = \frac{\vec{a}+\vec{b}+\vec{c}}{3}$. By symmetry, the other two medians divided in 2:1 yield the exact same point, proving they are concurrent.

Ans: $\sqrt{6}$ and $\sqrt{14}$. Reason: Adjacent sides are $\frac{\vec{d_1}+\vec{d_2}}{2}$ and $\frac{\vec{d_1}-\vec{d_2}}{2}$. $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, length $= \sqrt{4+1+1} = \sqrt{6}$. $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$, length $= \sqrt{1+4+9} = \sqrt{14}$.

Ans: $40$ Joules. Reason: Net force $\vec{F} = \vec{F_1} + \vec{F_2} = 7\hat{i} + 2\hat{j} - 4\hat{k}$. Displacement $\vec{d} = \vec{OQ} - \vec{OP} = 4\hat{i} + 2\hat{j} - 2\hat{k}$. Work $W = \vec{F} \cdot \vec{d} = 28 + 4 + 8 = 40$.

Ans: $\frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k}$. Reason: $\vec{a}\times\vec{b} = 15\hat{i} - 10\hat{j} + 30\hat{k}$. Magnitude = $\sqrt{225+100+900} = 35$. Unit vector = $\frac{1}{35}(15\hat{i} - 10\hat{j} + 30\hat{k})$. Simplify by dividing by 5.

Proof: Area = $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})| = \frac{1}{2} |\vec{b}\times\vec{c} - \vec{b}\times\vec{a} - \vec{a}\times\vec{c} + \vec{a}\times\vec{a}| = \frac{1}{2} |\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}|$.

Proof: $\vec{a}, \vec{b}, \vec{c}$ coplanar $\Rightarrow [\vec{a}\vec{b}\vec{c}] = 0$. The STP of the new vectors is $[\vec{a}\times\vec{b} \quad \vec{b}\times\vec{c} \quad \vec{c}\times\vec{a}] = [\vec{a}\vec{b}\vec{c}]^2 = 0^2 = 0$. Hence they are also coplanar.

Proof: Subtracting the given equations: $\vec{a}\times\vec{b} - \vec{a}\times\vec{c} = \vec{c}\times\vec{d} - \vec{b}\times\vec{d} \Rightarrow \vec{a}\times(\vec{b}-\vec{c}) = (\vec{c}-\vec{b})\times\vec{d} \Rightarrow \vec{a}\times(\vec{b}-\vec{c}) = -\vec{d}\times(\vec{b}-\vec{c})$. Moving terms: $(\vec{a}-\vec{d})\times(\vec{b}-\vec{c}) = \vec{0}$. Since cross product is zero and neither vector is zero, they must be parallel.

Ans: $6$ cubic units. Reason: Volume of new parallelepiped is $[\vec{a}\times\vec{b} \quad \vec{b}\times\vec{c} \quad \vec{c}\times\vec{a}] = [\vec{a}\vec{b}\vec{c}]^2 = 36$. Hence $[\vec{a}\vec{b}\vec{c}] = \pm 6$. The volume of the original parallelepiped is $|[\vec{a}\vec{b}\vec{c}]| = 6$.