Q.Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.

From $\vec a+\vec b+\vec c=\vec 0$ with $|\vec a|=3,|\vec b|=4,|\vec c|=5$, the angle between $\vec a$ and $\vec b$ works out to $\dfrac{\pi}{2}$.

1. From $\vec a+\vec b+\vec c=\vec0$, we get $\vec c=-(\vec a+\vec b)$.
2. Take magnitudes squared: $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2$.
3. Substitute the given magnitudes: $5^2=3^2+2\vec a\cdot\vec b+4^2$, i.e. $25=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b$.
4. So $2\vec a\cdot\vec b=0\Rightarrow \vec a\cdot\vec b=0$.
5. Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0$ and $|\vec a|,|\vec b|\neq0$, we must have $\cos\theta=0$.
6. Hence $\theta=\dfrac{\pi}{2}$.

The area of a quadrilateral in terms of its diagonal vectors $\vec d_1,\vec d_2$ is $\frac12|\vec d_1\times\vec d_2|$; here the diagonals are $\vec{PR}$ and $\vec{QS}$, so compute their cross product and halve its magnitude.

1. For quadrilateral PQRS, the diagonals are $PR$ and $QS$, and the standard vector-algebra result is $\text{Area}(PQRS) = \frac12\,|\vec{PR}\times\vec{QS}|.$
2. Given $\vec{PR}=2\vec i+\vec j+\vec k$ and $\vec{QS}=-\vec i+3\vec j+2\vec k$.
3. Compute the cross product: $\vec{PR}\times\vec{QS}=\begin{vmatrix}\vec i & \vec j & \vec k\\ 2 & 1 & 1\\ -1 & 3 & 2\end{vmatrix}$
4. $\vec i$-component: $(1)(2)-(1)(3) = 2-3=-1$.
5. $\vec j$-component: $-[(2)(2)-(1)(-1)] = -(4+1)=-5$.
6. $\vec k$-component: $(2)(3)-(1)(-1) = 6+1=7$.
7. So $\vec{PR}\times\vec{QS} = -\vec i-5\vec j+7\vec k$.
8. Magnitude: $\sqrt{(-1)^2+(-5)^2+7^2}=\sqrt{1+25+49}=\sqrt{75}=5\sqrt3$.
9. Area $= \dfrac12(5\sqrt3) = \dfrac{5\sqrt3}{2}$.
10. This matches option (c).

The vector $\vec i-\vec j$ has zero component along $\vec k$, so its projection on the $z$-axis is $0$.

1. The projection of a vector $\vec r=x\vec i+y\vec j+z\vec k$ on the $z$-axis equals $\vec r\cdot\vec k=z$.
2. Here $\vec r=\vec i-\vec j=1\vec i-1\vec j+0\vec k$.
3. So $\vec r\cdot\vec k=0$.
4. Options (b) $1$, (c) $-1$, (d) $2$ would arise from projecting on the $x$- or $y$-axis instead, or a sign slip.

Recovering the second side from the diagonal and given side, then taking the cross product, gives an area of $3\sqrt{30}$.

1. In a parallelogram with adjacent sides $\vec a$ and $\vec b$ from the same vertex, the diagonal from that vertex is $\vec d=\vec a+\vec b$.
2. Given side $\vec a=\vec i-3\vec j+4\vec k$ and diagonal $\vec d=3\vec i+\vec j-\vec k$, the other side is $\vec b=\vec d-\vec a=(3-1)\vec i+(1-(-3))\vec j+(-1-4)\vec k=2\vec i+4\vec j-5\vec k$.
3. Area of the parallelogram $=|\vec a\times\vec b|$.
4. Compute the cross product: $\vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\1&-3&4\\2&4&-5\end{vmatrix}$ $i$-component: $(-3)(-5)-(4)(4)=15-16=-1$. $j$-component: $-[(1)(-5)-(4)(2)]=-[-5-8]=13$. $k$-component: $(1)(4)-(-3)(2)=4+6=10$. So $\vec a\times\vec b=-\vec i+13\vec j+10\vec k$.
5. Magnitude: $\sqrt{(-1)^2+13^2+10^2}=\sqrt{1+169+100}=\sqrt{270}=\sqrt{9\cdot30}=3\sqrt{30}$.
6. This matches option (d); the other options come from miscomputing the second side or the cross product.