Q.Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

Concept and Intuition

For any vectors, $\big||\bar a|-|\bar b|\big|\le|\bar a-\bar b|\le|\bar a|+|\bar b|$ always holds (triangle inequality), so options claiming the reverse must be examined by direct computation rather than assumed.

Step-by-Step Solution

1. $\bar a-\bar b=(1-2,\,2-(-3),\,-3-(-5))=(-1,5,2)$, so $|\bar a-\bar b|=\sqrt{1+25+4}=\sqrt{30}$.
2. $\bar a+\bar b=(3,-1,-8)$, so $|\bar a+\bar b|=\sqrt{9+1+64}=\sqrt{74}$.
3. $|\bar a|=\sqrt{1+4+9}=\sqrt{14}\approx3.74$; $|\bar b|=\sqrt{4+9+25}=\sqrt{38}\approx6.16$.
4. Check (A): $\sqrt{30}\approx5.48$ vs $|\bar a|+|\bar b|\approx9.91$ — false.
5. Check (B): $\sqrt{30}\approx5.48$ vs $|\bar b|-|\bar a|\approx2.42$ — true. …

Concept and Intuition

Once $\bar a-\bar b+\bar c$ is written out component-wise, the magnitude condition becomes a simple quadratic equation in the unknown $y$ (since $y$ only appears in the $\bar j$-component here). Solving that quadratic gives two roots, and their difference is what's asked.

Step-by-Step Solution

1. $\bar a-\bar b+\bar c=(3\bar i+\bar j-2\bar k)-(-5\bar i+7\bar j)+(3\bar i+y\bar j)$.
2. $\bar i$-component: $3-(-5)+3=11$. $\bar j$-component: $1-7+y=y-6$. $\bar k$-component: $-2-0+0=-2$.
3. So the vector is $11\bar i+(y-6)\bar j-2\bar k$, with magnitude squared $=121+(y-6)^2+4=125+(y-6)^2$. …

Concept and Intuition

For unit vectors, the length of the difference vector is directly tied to the angle between them via the law-of-cosines-like expansion of the dot product, and the half-angle identity $1-\cos\theta=2\sin^2(\theta/2)$ finishes the connection.

Step-by-Step Solution

1. $|\vec a-\vec b|^2=(\vec a-\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2$.
2. Since $|\vec a|=|\vec b|=1$ and $\vec a\cdot\vec b=\cos\theta$: $|\vec a-\vec b|^2=1+1-2\cos\theta=2(1-\cos\theta)$.
3. Use $1-\cos\theta=2\sin^2(\theta/2)$: $|\vec a-\vec b|^2=4\sin^2(\theta/2)$. …

Concept and Intuition

For any two vectors, $|\bar a\pm\bar b|^2=|\bar a|^2+|\bar b|^2\pm2\bar a\cdot\bar b$. Given $|\bar a|,|\bar b|,|\bar a+\bar b|$, the middle (cross) term $\bar a\cdot\bar b$ is the only unknown — solve for it using the "+" identity, then plug it into the "−" identity. Here the numbers $3,4,5$ (a Pythagorean triple) are exactly what makes $\bar a\cdot\bar b$ come out to zero.

Step-by-Step Solution

1. Given $|\bar a|=3,\ |\bar b|=4,\ |\bar a+\bar b|=5$.
2. Expand: $|\bar a+\bar b|^2=|\bar a|^2+|\bar b|^2+2\bar a\cdot\bar b$.
3. Substitute: $25=9+16+2\bar a\cdot\bar b=25+2\bar a\cdot\bar b\Rightarrow 2\bar a\cdot\bar b=0\Rightarrow\bar a\cdot\bar b=0$. …

Concept and Intuition

For any two vectors, $||u|-|v||\le|u-v|\le|u|+|v|$. The right equality ($|u-v|=|u|+|v|$) happens when $u,v$ are anti-parallel; the left equality ($|u-v|=||u|-|v||$) happens when $u,v$ are parallel and point the same way (one is a nonnegative scalar multiple of the other in the same direction).

Step-by-Step Solution

1. Square both sides of $|u-v| = ||u|-|v||$: $|u-v|^2 = (|u|-|v|)^2$.
2. Expand LHS: $|u|^2 - 2u\cdot v + |v|^2$. Expand RHS: $|u|^2 - 2|u||v| + |v|^2$.
3. Cancel common terms: $-2u\cdot v = -2|u||v| \Rightarrow u\cdot v = |u||v|$.
4. Since $u\cdot v = |u||v|\cos\theta$, this forces $\cos\theta = 1$, i.e. $\theta = 0$.
5. $\theta=0$ means $u$ and $v$ point in the same direction.

Common Mistakes …

Concept and Intuition

In any square of side $s$, the diagonal has length $s\sqrt2$ (Pythagoras on two adjacent sides). So once we know the diagonal, the side follows immediately by dividing by $\sqrt2$ — this works the same in 3D as in 2D since the square itself is a planar figure.

Step-by-Step Solution

1. Diagonal $=\sqrt{(2-1)^2+((-3)-(-2))^2+(5-3)^2}=\sqrt{1^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt6$. …