Q.If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0.$ But the converse need not be true. Justify your answer with an example.

Concept and Intuition

For any two vectors, $|\vec a+\vec b| \le |\vec a|+|\vec b|$, with equality exactly when $\vec a$ and $\vec b$ are parallel and point the same way (one is a non-negative scalar multiple of the other). Squaring both sides confirms this: $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$ equals $(|\vec a|+|\vec b|)^2=|\vec a|^2+|\vec b|^2+2|\vec a||\vec b|$ only when $\vec a\cdot\vec b=|\vec a||\vec b|$, i.e. the angle between them is $0$.

Step-by-Step Solution

1. Given $\vec a=\vec i+x\vec j+\vec k$ and $\vec b=\vec i+\vec j+\vec k$.
2. The equality condition means $\vec a=\lambda\vec b$ for some scalar $\lambda\ge 0$.
3. Matching the $\vec i$-component: $1=\lambda\cdot 1 \Rightarrow \lambda=1$.
4. Matching the $\vec k$-component: $1=\lambda\cdot1$, consistent with $\lambda=1$.
5. Matching the $\vec j$-component: $x=\lambda\cdot1=1$. …

Concept and Intuition

Whenever three vectors sum to zero, dotting the relation with itself is the standard trick to relate the pairwise dot products to the (given) magnitudes, without needing any angle information.

Step-by-Step Solution

1. Start from $\bar a+\bar b+\bar c=\bar 0$.
2. Take the dot product of both sides with themselves: $(\bar a+\bar b+\bar c)\cdot(\bar a+\bar b+\bar c)=0$.
3. Expand: $|\bar a|^2+|\bar b|^2+|\bar c|^2+2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)=0$. …

Concept and Intuition

When three vectors sum to zero, each one is the negative of the sum of the other two — so its magnitude squared can be found from $|\vec u+\vec v|^2=|\vec u|^2+|\vec v|^2+2\vec u\cdot\vec v$.

Step-by-Step Solution

1. From $\vec a+\vec b+\vec c=\vec0$: $\vec c=-(\vec a+\vec b)$.
2. $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$. …

Concept and Intuition

The magnitude of a cross product involves $\sin\theta$ and the dot product involves $\cos\theta$, where $\theta$ is the angle between the vectors. Squaring and adding these naturally invokes $\sin^2\theta+\cos^2\theta=1$.

Step-by-Step Solution

1. $|u\times v| = |u||v|\sin\theta$, so $|u\times v|^2=|u|^2|v|^2\sin^2\theta$.
2. $u\cdot v = |u||v|\cos\theta$, so $|u\cdot v|^2=|u|^2|v|^2\cos^2\theta$. …

Concept and Intuition

$Q$ is just $|\bar a|^2$ split into its three squared components via dot products with $\bar i,\bar j,\bar k$. $P$ does the analogous thing with cross products — but crossing with a basis vector kills one component and swaps/negates the other two, so summing over all three basis vectors ends up counting each squared component of $\bar a$ twice.

Step-by-Step Solution

1. Let $\bar a = a_1\bar i+a_2\bar j+a_3\bar k$.
2. $Q = (\bar a\cdot\bar i)^2+(\bar a\cdot\bar j)^2+(\bar a\cdot\bar k)^2 = a_1^2+a_2^2+a_3^2$.
3. $\bar a\times\bar i = (a_1,a_2,a_3)\times(1,0,0) = (0,\ a_3,\ -a_2)$, so $|\bar a\times\bar i|^2 = a_2^2+a_3^2$.
4. $\bar a\times\bar j = (-a_3,\ 0,\ a_1)$, so $|\bar a\times\bar j|^2 = a_1^2+a_3^2$. …

Concept and Intuition

The dot product distributes over vector addition just like multiplication over addition in ordinary algebra, so $(\vec A+\vec B)\cdot\vec B$ splits cleanly into two known pieces.

Step-by-Step Solution

1. Expand: $(\vec A+\vec B)\cdot\vec B = \vec A\cdot\vec B + \vec B\cdot\vec B$.
2. We're given $\vec A\cdot\vec B=6$ and $\vec B\cdot\vec B=|\vec B|^2$.
3. So $6+|\vec B|^2=15 \Rightarrow |\vec B|^2=9$.
4. Since magnitude is non-negative: $|\vec B|=3$. …

Concept and Intuition

$|\vec a+\vec b|=25$ becomes a straightforward equation in $t$ once the vector sum is written component-wise — the $j$ and $k$ components are already fixed numbers, so only the $i$-component (which contains $t$) contributes a variable term to the magnitude.

Step-by-Step Solution

1. $\vec a=-2\vec i+9\vec j-6\vec k=(-2,9,-6)$, $\vec b=t\vec i-2\vec j+6\vec k=(t,-2,6)$.
2. $\vec a+\vec b=(t-2,\ 9-2,\ -6+6)=(t-2,\ 7,\ 0)$.
3. $|\vec a+\vec b|^2=(t-2)^2+7^2+0^2=(t-2)^2+49$. …

Concept and Intuition

$|\bar f\pm\bar g|^2 = |\bar f|^2+|\bar g|^2\pm2\bar f\cdot\bar g$ are the two 'parallelogram law' identities; knowing one combination lets you solve for the dot product, then use it in the other.

Step-by-Step Solution

1. $|\bar f-\bar g|^2 = |\bar f|^2+|\bar g|^2-2\bar f\cdot\bar g \Rightarrow 15^2 = 10^2+14^2-2\bar f\cdot\bar g$.
2. $225 = 100+196-2\bar f\cdot\bar g = 296-2\bar f\cdot\bar g$.
3. $2\bar f\cdot\bar g = 296-225 = 71 \Rightarrow \bar f\cdot\bar g = 35.5$.
4. $|\bar f+\bar g|^2 = |\bar f|^2+|\bar g|^2+2\bar f\cdot\bar g = 296+71 = 367$. …

Concept and Intuition

$a,b,c$ denote the standard triangle side lengths — $a=BC$ (opposite $A$), $b=CA$ (opposite $B$), $c=AB$ (opposite $C$). We are directly given the vectors $\overline{AB}$ and $\overline{AC}$, so $c=|\overline{AB}|$ and $b=|\overline{AC}|$ come immediately; the third side $\overline{BC}$ is obtained as $\overline{AC}-\overline{AB}$ (walk from $A$ to $C$ minus walk from $A$ to $B$).

Step-by-Step Solution

1. $c=|\overline{AB}|=\sqrt{2^2+(-3)^2+7^2}=\sqrt{4+9+49}=\sqrt{62}$.
2. $b=|\overline{AC}|=\sqrt{1^2+(-6)^2+5^2}=\sqrt{1+36+25}=\sqrt{62}$.
3. $\overline{BC}=\overline{AC}-\overline{AB}=(1-2,\,-6-(-3),\,5-7)=(-1,-3,-2)$.
4. $a=|\overline{BC}|=\sqrt{(-1)^2+(-3)^2+(-2)^2}=\sqrt{1+9+4}=\sqrt{14}$.
5. $a^2+b^2+c^2=14+62+62=138$.

Common Mistakes …

Concept and Intuition

Sums of squares of vectors like $(\bar a+\bar b-\bar c)^2$ over all cyclic permutations always reduce to a combination of $\sum|\cdot|^2$ and $\sum(\text{dot products})$; recognizing that the specific numeric value given forces $\bar a+\bar b+\bar c=0$ is the key simplification that makes the final vector combination tractable.

Step-by-Step Solution

1. Expand each square: $(\bar a+\bar b-\bar c)^2 = |\bar a|^2+|\bar b|^2+|\bar c|^2+2\bar a\cdot\bar b-2\bar a\cdot\bar c-2\bar b\cdot\bar c$, and similarly (cyclically) for the other two terms.
2. Summing all three, the cross terms telescope to $-2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)$, giving total $= 3(|\bar a|^2+|\bar b|^2+|\bar c|^2) - 2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)$.
3. Since $|\bar a|=|\bar b|=|\bar c|=\sqrt3$, each magnitude$^2=3$, so $3(9)=27$. Thus $27-2\Sigma=36\Rightarrow \Sigma=\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a=-4.5$.
4. Now compute $|\bar a+\bar b+\bar c|^2 = |\bar a|^2+|\bar b|^2+|\bar c|^2+2\Sigma = 9+2(-4.5)=0$. A vector with zero magnitude is the zero vector, so $\bar a+\bar b+\bar c=0$ — this is an exact consequence, not an assumption. …

Concept and Intuition

In any parallelogram (and a square is one), the diagonals bisect each other. For square $ABCD$ the diagonals are $AC$ and $BD$, so their midpoints coincide. This gives three scalar equations (one per coordinate) in the two unknowns $\lambda,\mu$ — enough to solve and then verify.

Step-by-Step Solution

1. Midpoint of $AC$: $\left(\dfrac{0+\lambda}{2},\dfrac{\lambda+5}{2},\dfrac{1+0}{2}\right)=\left(\dfrac\lambda2,\dfrac{\lambda+5}2,\dfrac12\right)$.
2. Midpoint of $BD$: $\left(\dfrac{\mu+\mu}{2},\dfrac{3+6}{2},\dfrac{-1+\mu}{2}\right)=\left(\mu,4.5,\dfrac{\mu-1}2\right)$.
3. Equate: $\dfrac\lambda2=\mu$; $\dfrac{\lambda+5}2=4.5\Rightarrow\lambda=4$; $\dfrac12=\dfrac{\mu-1}2\Rightarrow\mu=2$. Check first equation: $\lambda/2=2=\mu$ ✓.
4. So $A=(0,4,1),\,B=(2,3,-1),\,C=(4,5,0),\,D=(2,6,2)$. …

Concept and Intuition

In a general triangle, the orthocenter, centroid, and circumcenter are distinct points (lying on the Euler line). But in an equilateral triangle, by symmetry, ALL of these special points coincide at a single center. So "origin is the orthocenter" is equivalent here to "origin is the centroid."

Step-by-Step Solution

1. For an equilateral triangle with vertices having position vectors $\bar a,\bar b,\bar c$, the centroid's position vector is $G=\dfrac{\bar a+\bar b+\bar c}{3}$.
2. By the symmetry of an equilateral triangle, the centroid, orthocenter, and circumcenter are the same point.
3. We are told the origin is the orthocenter; since orthocenter = centroid here, the origin is also the centroid. …