Vector Algebra — JEE Main practice

1. Rearrange the given cross-product condition to show that $\vec{c}+\vec{d}$ is parallel to the vector $2\hat{i}+3\hat{j}+4\hat{k}$. Given $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$ So $(\vec{c}+\vec{d})\times(2\hat{i}+3\hat{j}+4\hat{k})=\vec{0} \Rightarrow \vec{c}+\vec{d}=\lambda(2\hat{i}+3\hat{j}+4\hat{k})$

2. Use the given magnitude $|\vec{c}+\vec{d}|=\sqrt{29}$ to find $\lambda$, noting that $|2\hat{i}+3\hat{j}+4\hat{k}|=\sqrt{29}$ as well. $\sqrt{29}=|\lambda|\sqrt{29} \Rightarrow \lambda=\pm1$ So $\vec{c}+\vec{d}=\pm(2\hat{i}+3\hat{j}+4\hat{k})$

3. Compute $(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$ for both signs to find $\lambda_1$ and $\lambda_2$. $(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})=\pm(-14+6+12)=\pm4$ So $\lambda_1=4,\quad \lambda_2=-4$

4. Apply the condition for the given second-degree equation to represent a circle: the coefficients of $x^2$ and $y^2$ must be equal, and the coefficient of $xy$ must be zero. Given the equation $K^2x^2+(K^2-5K+\lambda_1)xy+\left(3K+\frac{\lambda_2}{2}\right)y^2-8x+12y+\lambda_2=0$ So $K^2=3K+\frac{\lambda_2}{2}=3K-2 \quad\text{and}\quad K^2-5K+\lambda_1=0 \Rightarrow K^2-5K+4=0$

5. Solve each quadratic in $K$. $K^2-3K+2=0 \Rightarrow K=1,2$ $K^2-5K+4=0 \Rightarrow K=1,4$

6. Take the common value of $K$ satisfying both equations. $K=1$

Hence, the answer is Option D (K = 1).

1. Write the general point on the given line in terms of the parameter $\lambda$. Given $\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})$ So a general point is $(2\lambda-1,\ 3\lambda+3,\ 1-\lambda)$

2. Form the vector from the given point $(5,4,2)$ to this general point, since this vector must be perpendicular to the line's direction at the foot of the perpendicular. $\overrightarrow{PF}=(2\lambda-6)\hat{i}+(3\lambda-1)\hat{j}+(-\lambda-1)\hat{k}$

3. Use the perpendicularity condition ($\overrightarrow{PF}\cdot\text{direction}=0$) to solve for $\lambda$. Given $\overrightarrow{PF}\cdot(2\hat{i}+3\hat{j}-\hat{k})=0$ So $4\lambda-12+9\lambda-3+\lambda+1=0 \Rightarrow 14\lambda=14 \Rightarrow \lambda=1$

4. Substitute $\lambda=1$ to find the foot of the perpendicular $F=(\alpha,\beta,\gamma)$. $F\equiv(1,6,0)$

5. Use the projection formula (length $=|\vec{u}\cdot\vec{v}|/|\vec{v}|$) to find the projection of $\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ on the given vector. $\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}=\hat{i}+6\hat{j}$ $\left|\frac{(\hat{i}+6\hat{j})\cdot(6\hat{i}+2\hat{j}+3\hat{k})}{|6\hat{i}+2\hat{j}+3\hat{k}|}\right|=\frac{6+12}{7}=\frac{18}{7}$

Hence, the answer is Option A $\left(\frac{18}{7}\right)$.

1. Let $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ and use the given cross product relation $\vec{a}\times\vec{c}=\vec{b}$, expanded via the determinant method, to form a system of equations. Given $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k},\quad \vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ $\vec{a}\times\vec{c}=(2z-2y)\hat{i}+(z+2x)\hat{j}+(-y-2x)\hat{k}=8\hat{i}+7\hat{j}-3\hat{k}$

2. Equate the components to get three equations, and express $y$ and $z$ in terms of $x$. $2z-2y=8,\quad z+2x=7,\quad -y-2x=3$ So $y=3-2x,\quad z=7-2x$

3. Use the given condition $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$ to solve for $x$. Given $x+y+z=4$ So $x+(3-2x)+(7-2x)=4 \Rightarrow -3x+10=4 \Rightarrow x=2$

4. Find $y$ and $z$, then write $\vec{c}$ explicitly. $y=-1,\quad z=3 \Rightarrow \vec{c}=2\hat{i}-\hat{j}+3\hat{k}$

5. Compute $\vec{a}+\vec{c}$ and its squared magnitude. $\vec{a}+\vec{c}=\hat{i}+\hat{j}+5\hat{k}$ $|\vec{a}+\vec{c}|^2=1+1+25=27$

Hence, the answer is Option C (27).