Scalar or Dot product of two vectors and its applications Questions in English

(B) Given $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$, we can write $(\vec{r} - \vec{c}) \times \vec{b} = 0$. This implies $\vec{r} - \vec{c} = t\vec{b}$ for some scalar $t$, so $\vec{r} = \vec{c} + t\vec{b}$.
Substituting $\vec{r}$ into $\vec{r} \cdot \vec{a} = 0$, we get $(\vec{c} + t\vec{b}) \cdot \vec{a} = 0$, which means $\vec{c} \cdot \vec{a} + t(\vec{b} \cdot \vec{a}) = 0$.
Calculate $\vec{c} \cdot \vec{a} = (4)(2) + (-3)(0) + (7)(1) = 8 + 0 + 7 = 15$.
Calculate $\vec{b} \cdot \vec{a} = (1)(2) + (1)(0) + (1)(1) = 2 + 0 + 1 = 3$.
Thus, $15 + 3t = 0$, which gives $t = -5$.
Now, $\vec{r} = \vec{c} - 5\vec{b} = (4\hat{i} - 3\hat{j} + 7\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - 8\hat{j} + 2\hat{k}$.
Finally, $\vec{r} \cdot \vec{c} = (-1)(4) + (-8)(-3) + (2)(7) = -4 + 24 + 14 = 34$.

(A) Step $1$: Since the lines intersect at $(2, 4, -4)$, the point must satisfy both equations.
Step $2$: For the first line, $\vec{a} + \lambda\vec{b} = (2, 4, -4)$. For the second line, $\vec{b} + \mu\vec{a} = (2, 4, -4)$.
Step $3$: Subtracting the two equations: $(\vec{a} - \vec{b}) + (\lambda\vec{b} - \mu\vec{a}) = 0$. This implies $\vec{a}(1 - \mu) = \vec{b}(1 - \lambda)$.
Step $4$: Since the lines intersect at the same point, $\vec{a} + \lambda\vec{b} = \vec{b} + \mu\vec{a} \implies \vec{a}(1 - \mu) = \vec{b}(1 - \lambda)$. If $\vec{a}$ and $\vec{b}$ are not collinear, then $1-\mu = 0$ and $1-\lambda = 0$, so $\lambda = 1$ and $\mu = 1$.
Step $5$: Substituting $\lambda = 1$ into the first equation: $\vec{a} + \vec{b} = (2, 4, -4)$.
Step $6$: We are given $|\vec{a} - \vec{b}| = 4$. Squaring both sides: $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 16$.
Step $7$: From $\vec{a} + \vec{b} = (2, 4, -4)$, we have $|\vec{a} + \vec{b}|^2 = 2^2 + 4^2 + (-4)^2 = 4 + 16 + 16 = 36$. So $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 36$.
Step $8$: Subtracting the two equations: $(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}) - (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) = 36 - 16 \implies 4\vec{a} \cdot \vec{b} = 20 \implies \vec{a} \cdot \vec{b} = 5$.

(D) Given $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.
Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.
Using the property $|\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2(\vec{u} \cdot \vec{v})$, we have $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
This simplifies to $4(\vec{a} \cdot \vec{b}) = 0$, which implies $\vec{a} \cdot \vec{b} = 0$.
Calculating the dot product: $(2\hat{i} + 2\hat{j} - \hat{k}) \cdot (\alpha\hat{i} + \beta\hat{j} + 2\hat{k}) = 0$.
$2\alpha + 2\beta - 2 = 0$.
Dividing by $2$, we get $\alpha + \beta - 1 = 0$, so $\alpha + \beta = 1$.

(B) Two vectors are orthogonal if their dot product is zero, i.e., $\vec{a} \cdot \vec{b} = 0$.
Given $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Calculating the dot product: $(2)(1) + (\lambda)(2) + (1)(3) = 0$.
$2 + 2\lambda + 3 = 0$.
$2\lambda + 5 = 0$.
$2\lambda = -5$.
$\lambda = -\frac{5}{2}$.

(B) Step $1$: Find the vectors $\vec{AB}$ and $\vec{BC}$.
$\vec{AB} = (4-2)\hat{i} + (a-4)\hat{j} + (9-3)\hat{k} = 2\hat{i} + (a-4)\hat{j} + 6\hat{k}$.
$\vec{BC} = (10-4)\hat{i} + (-1-a)\hat{j} + (7-9)\hat{k} = 6\hat{i} - (1+a)\hat{j} - 2\hat{k}$.
Step $2$: Since $\angle B = 90^\circ$, the dot product $\vec{AB} \cdot \vec{BC} = 0$.
Step $3$: Calculate the dot product: $2(6) + (a-4)(-(1+a)) + 6(-2) = 0$.
Step $4$: Simplify the equation: $12 - (a-4)(a+1) - 12 = 0$.
Step $5$: Solve for $a$: $-(a-4)(a+1) = 0$, which gives $(a-4)(a+1) = 0$.
Therefore, $a = 4$ or $a = -1$.