Let vecc$\vec{c}$ and vecd$\vec{d}$ be vectors such that |vecc+vecd|=sqrt29$|\vec{c}+\vec{d}|=\sqrt{29}$ and vecctimes(2hati+3hatj+4hatk)=(2hati+3hatj+4hatk)timesvecd$\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$ . If lambda_1,lambda_2(lambda_1>lambda_2)$\lambda_{1},\lambda_{2}(\lambda_{1}>\lambda_{2})$ are the possible values of (vecc+vecd).(-7hati+2hatj+3hatk)$(\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k})$ , then the equation K^2x^2+(K^2-5K+lambda_1)xy+left(3K+fraclambda_22right)y^2-8x+12y+lambda_2=0$K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+\left(3K+\frac{\lambda_{2}}{2}\right)y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for k equal to:
A.4
B.1
C.-1
D.2
Solution & Explanation
### Related Formula
vecu times vecv = -(vecv times vecu)$$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$
General equation of a circle requires coefficient of x^2$x^2$ = coefficient of y^2$y^2$ and coefficient of xy$xy$ = 0$0$.
### Core Logic
Given vecc times (2hati + 3hatj + 4hatk) = (2hati + 3hatj + 4hatk) times vecd$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$
Rearranging:
vecc times (2hati + 3hatj + 4hatk) + vecd times (2hati + 3hatj + 4hatk) = vec0$$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) + \vec{d} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$(vecc + vecd) times (2hati + 3hatj + 4hatk) = vec0$$(\vec{c} + \vec{d}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$
This implies vecc + vecd$\vec{c} + \vec{d}$ is parallel to 2hati + 3hatj + 4hatk$2\hat{i} + 3\hat{j} + 4\hat{k}$.
Let vecc + vecd = lambda (2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$.
### Step 1: Find the scaling factor
Given |vecc + vecd| = sqrt29$|\vec{c} + \vec{d}| = \sqrt{29}$|lambda| sqrt2^2 + 3^2 + 4^2 = sqrt29$$|\lambda| \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$$|lambda| sqrt4 + 9 + 16 = sqrt29$$|\lambda| \sqrt{4 + 9 + 16} = \sqrt{29}$$|lambda| sqrt29 = sqrt29 Rightarrow lambda = pm 1$$|\lambda| \sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$$
Therefore, vecc + vecd = pm(2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$.
### Step 2: Evaluate the dot product limits
We need values of (vecc + vecd) cdot (-7hati + 2hatj + 3hatk)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$.
Dot product = lambda(2(-7) + 3(2) + 4(3)) = lambda(-14 + 6 + 12) = 4lambda$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$.
Since lambda = pm 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$.
Given lambda_1 > lambda_2$\lambda_1 > \lambda_2$, we have lambda_1 = 4$\lambda_1 = 4$ and lambda_2 = -4$\lambda_2 = -4$.
### Step 3: Apply circle constraints
Substitute lambda_1$\lambda_1$ and lambda_2$\lambda_2$ into the conic equation:
K^2 x^2 + (K^2 - 5K + 4)xy + left(3K - 2right)y^2 - 8x + 12y - 4 = 0$$K^2 x^2 + (K^2 - 5K + 4)xy + \left(3K - 2\right)y^2 - 8x + 12y - 4 = 0$$
For this to represent a circle:
1) Coefficient of xy$xy$ must be 0$0$:
K^2 - 5K + 4 = 0 Rightarrow K = 1, 4$$K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$$
2) Coefficient of x^2$x^2$ must equal coefficient of y^2$y^2$:
K^2 = 3K - 2 Rightarrow K^2 - 3K + 2 = 0 Rightarrow K = 1, 2$$K^2 = 3K - 2 \Rightarrow K^2 - 3K + 2 = 0 \Rightarrow K = 1, 2$$
The common value satisfying both conditions is K = 1$K = 1$.
### Pattern Recognition
Cross-product equations mapping to X times A = -Y times A$X \times A = -Y \times A$ perfectly factor out to (X+Y) times A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Class 11 Maths: Conic Sections
Q68jee_main_2025_02_april_eveningProperties of Vectors
Let veca = 2hati - 3hatj + hatk$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, vecb = 3hati + 2hatj + 5hatk$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$ and a vector vecc$\vec{c}$ be such that (veca - vecc) times vecb = -18hati - 3hatj + 12hatk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and veca cdot vecc = 3$\vec{a} \cdot \vec{c} = 3$. If vecb times vecc = vecd$\vec{b} \times \vec{c} = \vec{d}$, then |veca cdot vecd|$|\vec{a} \cdot \vec{d}|$ is equal to:
A. 18
B. 12
C. 9
D. 15
Solution
### Related Formula
textVector Cross product distributes over subtraction: (veca - vecc) times vecb = veca times vecb - vecc times vecb$$\text{Vector Cross product distributes over subtraction: } (\vec{a} - \vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$$textScalar Triple Product cyclic identity: veca cdot (vecb times vecc) = (veca times vecb) cdot vecc$$\text{Scalar Triple Product cyclic identity: } \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$$textAntisymmetry: vecc times vecb = - vecb times vecc$$\text{Antisymmetry: } \vec{c} \times \vec{b} = - \vec{b} \times \vec{c}$$
### Core Logic
Instead of solving for the individual coordinates of vector vecc$\vec{c}$, we apply vector algebraic identities to compute the target scalar triple product directly.
### Step 1: Expand and rewrite the cross product
Given (veca - vecc) times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$:
veca times vecb - vecc times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$veca times vecb + vecb times vecc = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (veca times vecb) quad text--- (1)$$\vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (\vec{a} \times \vec{b}) \quad \text{--- (1)}$$
### Step 2: Calculate a x b
Evaluate the cross product:
veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ 2 & -3 & 1 \\ 3 & 2 & 5 endvmatrix$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$$veca times vecb = hatmathrmi(-15 - 2) - hatmathrmj(10 - 3) + hatmathrmk(4 - (-9)) = -17hatmathrmi - 7hatmathrmj + 13hatmathrmk$$\vec{a} \times \vec{b} = \hat{\mathrm{i}}(-15 - 2) - \hat{\mathrm{j}}(10 - 3) + \hat{\mathrm{k}}(4 - (-9)) = -17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}}$$
### Step 3: Solve for the vector d
Substitute veca times vecb$\vec{a} \times \vec{b}$ back into equation (1):
vecd = vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (-17hatmathrmi - 7hatmathrmj + 13hatmathrmk)$$\vec{d} = \vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (-17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}})$$vecd = -hatmathrmi + 4hatmathrmj - hatmathrmk$$\vec{d} = -\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}}$$
### Step 4: Compute the final dot product
Now compute the requested dot product:
veca cdot vecd = (2hatmathrmi - 3hatmathrmj + hatmathrmk) cdot (-hatmathrmi + 4hatmathrmj - hatmathrmk)$$\vec{a} \cdot \vec{d} = (2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}) \cdot (-\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}})$$veca cdot vecd = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$\vec{a} \cdot \vec{d} = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$left| veca cdot vecd right| = 15$$\left| \vec{a} \cdot \vec{d} \right| = 15$$
### Pattern Recognition
Scalar triple product shortcut: Recognizing that veca cdot vecd = veca cdot (vecb times vecc) = [ veca \, vecb \, vecc ]$\vec{a} \cdot \vec{d} = \vec{a} \cdot (\vec{b} \times \vec{c}) = [ \vec{a} \, \vec{b} \, \vec{c} ]$ allows you to find the scalar value through simple determinants and linear equations instead of solving for the vector components.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Qjee_main_2025_02_april_morningVector Projections
If veca$\vec{a}$ is a nonzero vector such that its projections on the vectors 2hatmathbfi - hatmathbfj + 2hatmathbfk$2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}}$, hatmathbfi + 2hatmathbfj - 2hatmathbfk$\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$ and hatmathbfk$\hat{\mathbf{k}}$ are equal, then a unit vector along veca$\vec{a}$ is:
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