Let a = i + 2 j + 3 k$\vec{\mathbf{a}} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$ , b = 3 i + j - k$\vec{\mathbf{b}} = 3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$ and \|\vec{\mathbf{c}}\| be three vectors such that c$\vec{\mathbf{c}}$ is coplanar with a$\vec{\mathbf{a}}$ and b$\vec{\mathbf{b}}$ . If the vector c$\vec{\mathbf{c}}$ is perpendicular to b$\vec{\mathbf{b}}$ and a · c = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$ , then | c|$\|\vec{\mathbf{c}}\|$ is equal to :

Related Formula

Cos angle with z-axis: θ = a · k| a|$\cos\theta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|}$. Obtuse angle condition: a · b < 0$\vec{a} \cdot \vec{b} < 0$.

Core Logic

• Angle with z-axis:

Since (π)/(6) < θ < (π)/(2) 0 < θ < √(3)2$\frac{\pi}{6} < \theta < \frac{\pi}{2} \implies 0 < \cos\theta < \frac{\sqrt{3}}{2}$:

Given λ > 0$\lambda > 0$, we get λ in (0, 3)$\lambda \in (0, 3)$.

Step 1: Obtuse Angle Condition

• a · b < 0$\vec{a} \cdot \vec{b} < 0$:

Step 2: Combine Intervals

Combining results: λ in (0, 3) - 2$\lambda \in (0, 3) - \{2\}$. Here α = 0, β = 3, γ = 2 α + β + γ = 5$\alpha = 0, \beta = 3, \gamma = 2 \implies \alpha + \beta + \gamma = 5$.

Pattern Recognition

Intersect dot product negativity condition with direction cosine angle inequality.

Chapter Mix

Class 12 Maths: Vector Algebra

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Core Logic

Given c = a × b$\vec{c} = \vec{a} \times \vec{b}$ and d = c × a$\vec{d} = \vec{c} \times \vec{a}$. Substitute c$\vec{c}$ into the expression for d$\vec{d}$:

Using the vector triple product expansion:

Step 1: Calculate Magnitudes and Dot Products

For a = - i + j + 2 k$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$ and b = i - j - 3 k$\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$:

Step 2: Express vector d

Substitute values into d$\vec{d}$:

Step 3: Evaluate Final Expression

Now compute ( a - b) · d$(\vec{a} - \vec{b}) \cdot \vec{d}$:

Substitute the known values:

Pattern Recognition

Instead of computing cross products sequentially (which is tedious and error-prone), immediately expand nested cross products using the standard vector triple product identity BAC-CAB$BAC-CAB$.

Chapter Mix

Class 12 Maths: Vector Algebra

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Core Logic

First, find v = a × b$\vec{v} = \vec{a} \times \vec{b}$.

Given v· c = 11$\vec{v}\cdot\vec{c} = 11$ and c = λ i + j + k$\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$:

Step 1: Projection Calculation

Now, c = i + j + k$\vec{c} = \hat{i} + \hat{j} + \hat{k}$. The length of the projection of b$\vec{b}$ on c$\vec{c}$ is p$p$:

Calculate 9p²$9p^2$:

Pattern Recognition

Standard sequence: compute cross product to find normal vector v$\vec{v}$, take dot product to deduce missing parameter λ$\lambda$, then substitute into scalar projection formula.

Chapter Mix

Class 12 Maths: Vector Algebra

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Core Logic

This implies that ( b - 2 c)$(\vec{b} - 2\vec{c})$ is collinear with a$\vec{a}$. So, b - 2 c = λ a$\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar λ$\lambda$.

Step 1: Finding Lambda

Square both sides of the relation:

Given | a| = 1, | b| = 4, | c| = 2$|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$ and the angle between b$\vec{b}$ and c$\vec{c}$ is 60^°$60^\circ$:

Substitute the values:

λ = ± 4$\lambda = \pm 4$

Step 2: Calculating Dot Product

We have b - 2 c = ± 4 a$\vec{b} - 2\vec{c} = \pm 4\vec{a}$. Take the dot product with c$\vec{c}$ on both sides:

Pattern Recognition

A cross-product equation structured as a×( X)=0$\vec{a}\times(\vec{X})=0$ immediately gives X = λ a$\vec{X} = \lambda\vec{a}$. Expanding the magnitude squared is the standard method to expose the dot products and solve for λ$\lambda$.

Chapter Mix

Class 12 Maths: Vector Algebra

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Step 1: Finding Vector c

However, evaluating carefully from the pdf steps, wait, 1 × -2$1 \times -2$ is -2$-2$, so - j(0 - (-2))$-\hat{j}(0 - (-2))$ is -2 j$-2\hat{j}$. The PDF indicates c = 2 i - 2 j + k$\vec{c} = 2\hat{i} - 2\hat{j} + \hat{k}$ (magnitude is 3$3$). Let's use | c| = √(4 + 4 + 1) = 3$|\vec{c}| = \sqrt{4 + 4 + 1} = 3$.

Step 2: Finding magnitude of d

Step 3: Calculating dot product

Given | d - a| = √(11)$|\vec{d} - \vec{a}| = \sqrt{11}$. Squaring both sides:

We have | a|² = 2² + 1² + (-2)² = 9$|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 9$.

Pattern Recognition

Isolate the unknown dot product by exploiting vector magnitude identities (difference square expansion). The cross-product magnitude uniquely isolates the remaining magnitude variable.

Chapter Mix

Class 12 Maths: Vector Algebra