Q.Compute the magnitude of the following vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$; $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$; $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$

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. $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$.
3. Substitute $|\vec a|=|\vec b|=2$, $\vec a\cdot\vec b=2$: $|\vec c|^2=4+4+2(2)=12$.
4. $|\vec c|=\sqrt{12}=2\sqrt3$.

Common Mistakes

• Forgetting the cross term $2\vec a\cdot\vec b$ when squaring the vector sum.

ANSWER: B

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.
5. So $\bar c = -(\bar a+\bar b)$. Substitute into the target: $2\bar a-3\bar b+2\bar c = 2\bar a-3\bar b+2(-\bar a-\bar b) = 2\bar a-3\bar b-2\bar a-2\bar b = -5\bar b$.
6. $|2\bar a-3\bar b+2\bar c|^2 = |-5\bar b|^2 = 25|\bar b|^2 = 25(3) = 75$.

Common Mistakes

• Trying to solve for individual dot products $\bar a\cdot\bar b,\bar b\cdot\bar c,\bar c\cdot\bar a$ separately (impossible from the given data alone) instead of noticing the sum forces $\bar a+\bar b+\bar c=0$, which sidesteps needing them individually.
• Sign slip when substituting $\bar c=-(\bar a+\bar b)$ into the target combination.

ANSWER: D

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: $|\bar a|^2+|\bar b|^2+|\bar c|^2+2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)=0$.
4. Substitute $|\bar a|=1,|\bar b|=3,|\bar c|=4$: $1+9+16+2S=0$, i.e. $26+2S=0$.
5. Solve: $S=-13$.

Common Mistakes

• Forgetting the factor of 2 in front of the cross terms when expanding the square.
• Sign errors when moving $26$ to the other side.

ANSWER: C

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

• Computing $\overline{BC}$ as $\overline{AB}-\overline{AC}$ instead of $\overline{AC}-\overline{AB}$ (sign doesn't matter for the magnitude squared here, but the direction convention matters in other problems — good habit to get right).
• Mislabeling which side is $a$, $b$, or $c$ (doesn't affect this particular sum, but is a common source of error in related problems).

ANSWER: A

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: $x=\lambda\cdot1=1$.
6. So $x=1$, and indeed $\vec a=\vec b=\vec i+\vec j+\vec k$, giving $|\vec a+\vec b|=2\sqrt3=|\vec a|+|\vec b|$. Confirmed.

Common Mistakes

• Squaring the equation carelessly and concluding only $\vec a\cdot\vec b = |\vec a||\vec b|$ without translating that into the componentwise parallel condition.
• Allowing $\lambda<0$, which would actually make $|\vec a+\vec b|<|\vec a|+|\vec b|$, not equal.

ANSWER: A

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: $|\vec B|=3$.

(Note: $\vec A=\hat i+2\hat j$ is not actually needed for this computation — it's provided context/consistency for the constraint but the answer follows purely from the two dot-product values given.)

Common Mistakes

• Trying to solve for the individual components of $\vec B$ instead of directly isolating $|\vec B|^2$.
• Forgetting $\vec B\cdot\vec B=|\vec B|^2$ (not $|\vec B|$).

ANSWER: D

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 $|u\cdot v|^2=|u|^2|v|^2\cos^2\theta$.
3. Sum: $|u|^2|v|^2\sin^2\theta + |u|^2|v|^2\cos^2\theta = |u|^2|v|^2(\sin^2\theta+\cos^2\theta) = |u|^2|v|^2$.

Common Mistakes

• Confusing this with the vector identity $|u\times v|^2 = |u|^2|v|^2 - (u\cdot v)^2$, which is the same fact rearranged — either form leads to the same answer here.

ANSWER: C

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 $|\bar a\times\bar j|^2 = a_1^2+a_3^2$.
5. $\bar a\times\bar k = (a_2,\ -a_1,\ 0)$, so $|\bar a\times\bar k|^2 = a_1^2+a_2^2$.
6. $P = (a_2^2+a_3^2)+(a_1^2+a_3^2)+(a_1^2+a_2^2) = 2(a_1^2+a_2^2+a_3^2) = 2Q$.

Common Mistakes

• Sign errors while computing each 2-component cross product.
• Forgetting to add all three cross-product magnitudes before comparing to $Q$.

ANSWER: B

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. $|\bar f+\bar g|^2 = |\bar f|^2+|\bar g|^2+2\bar f\cdot\bar g = 296+71 = 367$.
5. $|\bar f+\bar g| = \sqrt{367}$.

Common Mistakes

• Sign error when expanding $|\bar f-\bar g|^2$ (using $+2\bar f\cdot\bar g$ instead of $-2\bar f\cdot\bar g$).
• Trying to compute $|\bar f+\bar g|$ as $|\bar f|+|\bar g|$ directly (only true when vectors are parallel).

ANSWER: B

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. $|\vec a+\vec b|^2=(t-2)^2+7^2+0^2=(t-2)^2+49$.
4. Given $|\vec a+\vec b|=25$: $(t-2)^2+49=625\Rightarrow(t-2)^2=576\Rightarrow t-2=\pm24$.
5. So $t=26$ or $t=-22$. Sum of the values of $t$: $26+(-22)=4$.

Common Mistakes

• Forgetting there are two roots ($\pm24$) and reporting only one value of $t$.
• Arithmetic slip in $625-49=576$ or $\sqrt{576}=24$.

ANSWER: C

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 $A=(0,4,1),\,B=(2,3,-1),\,C=(4,5,0),\,D=(2,6,2)$.
5. Side vectors: $\vec{AB}=(2,-1,-2)$, $|\vec{AB}|=\sqrt{4+1+4}=3$; $\vec{BC}=(2,2,1)$, $|\vec{BC}|=3$; similarly $|\vec{CD}|=|\vec{DA}|=3$.
6. Check perpendicularity: $\vec{AB}\cdot\vec{BC}=4-2-2=0$, confirming a right angle — so $ABCD$ is genuinely a square of side $3$.

Common Mistakes

• Assuming equal side lengths alone prove a square — a rhombus also has equal sides; the right-angle check (dot product $=0$) is essential.
• Sign/order slip when writing the vertices in the stated cyclic order $A\to B\to C\to D$.

ANSWER: C

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.
4. So $\dfrac{\bar a+\bar b+\bar c}{3} = \bar 0 \Rightarrow \bar a+\bar b+\bar c=\bar 0 \Rightarrow \bar a+\bar b=-\bar c$.