Let a = i + 2 j + k$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and b = 2 i + j - k$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let c$\hat{c}$ be a unit vector in the plane of the vectors a$\vec{a}$ and b$\vec{b}$ and be perpendicular to a$\vec{a}$. Then such a vector c$\hat{c}$ is:

Related Formula

Cross product for vector perpendicular direction alignment:

Angle projection formula:

Core Logic

Compute cross product of b$\vec{b}$ and c$\vec{c}$:

Hence, unit vector a$\hat{a}$ matches form:

Step 1: Isolate Core Angle Direction

Check conditions against vector v = i + j + k$\vec{v} = \hat{i} + \hat{j} + \hat{k}$: Using a = i - j - k√(3)$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$:

This confirms the direction for a$\hat{a}$.

Step 2: Solve for Unknown Scalar Variable

Now compute angle with vector i + α j + k$\hat{i} + \alpha\hat{j} + \hat{k}$ for θ = (π)/(3)$\theta = \frac{\pi}{3}$:

Since left hand side is positive, α$\alpha$ must be strictly negative. Squaring both sides:

Since α < 0$\alpha < 0$, α = -√(6)$\alpha = -\sqrt{6}$.

Pattern Recognition

Keep strict track of signs when dealing with algebra containing square roots. Checking value constraints early on allows you to drop phantom positive/negative branches seamlessly.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Related Formula

Vector magnitude expansion identity:

Core Logic

Calculate the reference vector values: a = i + j + k | a|² = 3$\vec{a} = \hat{i} + \hat{j} + \hat{k} \implies |\vec{a}|^2 = 3$ b = 2 i + 2 j + k$\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$ d = a × b = - i + j | d|² = 2$\vec{d} = \vec{a} \times \vec{b} = -\hat{i} + \hat{j} \implies |\vec{d}|^2 = 2$

Expanding | c - 2 a|² = 8$|\vec{c} - 2\vec{a}|^2 = 8$:

Substituting | a|² = 3$|\vec{a}|^2 = 3$ and a · c = | c|$\vec{a} \cdot \vec{c} = |\vec{c}|$:

Step 1: Evaluating the Cross Product Vector Component

Using the given angle (π)/(4)$\frac{\pi}{4}$ between d$\vec{d}$ and c$\vec{c}$:

Step 2: Solving for the Vector Dot Product

Expanding using standard vector identities yields b · c = (8)/(3)$\vec{b} \cdot \vec{c} = \frac{8}{3}$.

Substituting this back into the target expression:

Pattern Recognition

Recognizing a perfect square trinomial (x² - 4x + 4 = 0$x^2 - 4x + 4 = 0$) instantly isolates the unknown vector magnitude.

Chapter Mix

Class 12 Maths: Vectors

Related Formula

Vector cross distribution property:

Core Logic

Gather cross products to solve for collineation lines [cite: 1472, 1473]: ( a - b) × d = 0 d = t( a - b)$$(\vec{a} - \vec{b}) \times \hat{d} = 0 \implies \hat{d} = t(\vec{a} - \vec{b})$$ [cite: 1473, 1474]

Compute the baseline difference vector [cite: 1475]: a - b = (1-3) i + (1-2) j + (1 - (-1)) k = -2 i - j + 2 k$$\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1 - (-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$$ [cite: 1475] d = t(-2 i - j + 2 k)$$\hat{d} = t(-2\hat{i} - \hat{j} + 2\hat{k})$$ [cite: 1475]

Since d$\hat{d}$ is a unit vector[cite: 1476]: |t| · √((-2)² + (-1)² + 2²) = 1 3|t| = 1 |t| = (1)/(3)$$|t| \cdot \sqrt{(-2)^2 + (-1)^2 + 2^2} = 1 \implies 3|t| = 1 \implies |t| = \frac{1}{3}$$ [cite: 1476, 1479]

Step 1: Applying orthogonality constraints

Using orthogonal information given for vectors c$\vec{c}$ and a$\vec{a}$ [cite: 1482]: c · a = 0 (0)(1) + λ(1) + μ(1) = 0 μ = -λ$$\vec{c} \cdot \vec{a} = 0 \implies (0)(1) + \lambda(1) + \mu(1) = 0 \implies \mu = -\lambda$$ [cite: 1483, 1484] c = λ( j - k) | c|² = 2λ²$$\vec{c} = \lambda(\hat{j} - \hat{k}) \implies |\vec{c}|^2 = 2\lambda^2$$ [cite: 1485]

Use product condition c · d = 1$\vec{c} \cdot \hat{d} = 1$ to isolate scalar values [cite: 1486]: t(-2 i - j + 2 k) · λ( j - k) = 1$$t(-2\hat{i} - \hat{j} + 2\hat{k}) \cdot \lambda(\hat{j} - \hat{k}) = 1$$ [cite: 1487] tλ(-1 - 2) = 1 -3tλ = 1 tλ = -(1)/(3)$$t\lambda(-1 - 2) = 1 \implies -3t\lambda = 1 \implies t\lambda = -\frac{1}{3}$$ [cite: 1488]

Since |t|² = (1)/(9)$|t|^2 = \frac{1}{9}$, squaring components yields [cite: 1488]: λ² = 1$\lambda^2 = 1$ [cite: 1488]

Step 2: Vector magnitude resolution

Expand target expression using standard inner dot product expansions [cite: 1488]: |3λ d + μ c|² = 9λ²| d|² + μ²| c|² + 6λμ( d · c)$$|3\lambda \hat{d} + \mu \vec{c}|^2 = 9\lambda^2|\hat{d}|^2 + \mu^2|\vec{c}|^2 + 6\lambda\mu(\hat{d} \cdot \vec{c})$$ [cite: 1488] Substitute values evaluated throughout sections [cite: 1488, 1489]:

= 9 + 2λ⁴ - 6λ² = 9 + 2(1) - 6(1) = 5$$= 9 + 2\lambda^4 - 6\lambda^2 = 9 + 2(1) - 6(1) = 5$$ [cite: 1489, 1490]

Pattern Recognition

Translating vector cross equalities directly into linear scale parameter multipliers prevents manual determinant expansions, leaving clean system variables behind.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Core Logic

Let the vertices of the triangle be A$A$, B$B$, and C$C$. The vectors representing the side paths are:

Notice that AB + CA = (2+1) i + (-1-3) j + (1-5) k = 3 i - 4 j - 4 k = CB$\overline{AB} + \overline{CA} = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k} = 3\hat{i} - 4\hat{j} - 4\hat{k} = \overline{CB}$. This structurally validates vector addition rules.

Step 1: Finding Position Vectors relative to A

Let's set vertex A$A$ as the origin origin point (Position vector A = 0$\vec{A} = \vec{0}$):

• Position vector of B$B$: B = 2 i - j + k$\vec{B} = 2\hat{i} - \hat{j} + \hat{k}$
• Position vector of C$C$: Since CA = A - C = - C C = - i + 3 j + 5hatk$\overline{CA} = \vec{A} - \vec{C} = -\vec{C} \implies \vec{C} = -\hat{i} + 3\hat{j} + 5hat{k}$

Now, calculate the position vector of the centroid G$G$:

Step 2: Calculating Squared Lengths to the Centroid

Let's find each individual vector distance block:

• AG = G - A = (1)/(3)( i + 2 j + 6 k) | AG|² = (1)/(9)(1² + 2² + 6²) = (41)/(9)$\overline{AG} = \vec{G} - \vec{A} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k}) \implies |\overline{AG}|^2 = \frac{1}{9}(1^2 + 2^2 + 6^2) = \frac{41}{9}$
• BG = G - B = ((1)/(3)-2) i + ((2)/(3)+1) j + (2-1) k = -(5)/(3) i + (5)/(3) j + 1 k$\overline{BG} = \vec{G} - \vec{B} = \left(\frac{1}{3}-2\right)\hat{i} + \left(\frac{2}{3}+1\right)\hat{j} + \left(2-1\right)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + 1\hat{k}$

• CG = G - C = ((1)/(3)+1) i + ((2)/(3)-3) j + (2-5) k = (4)/(3) i - (7)/(3) j - 3 k$\overline{CG} = \vec{G} - \vec{C} = \left(\frac{1}{3}+1\right)\hat{i} + \left(\frac{2}{3}-3\right)\hat{j} + \left(2-5\right)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$

Step 3: Final Targeted Evaluation

Summing the squared values and multiplying by 6$6$:

Pattern Recognition

Setting one vector node as the origin point (A = 0$\vec{A} = \vec{0}$) heavily dampens intermediate coordinate math steps, avoiding dealing with an absolute baseline origin orientation.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Related Formula

By orthogonal vector decomposition (Pythagorean property):

Core Logic

Compute λ$\lambda$ using dot product formula:

Step 1: Solve for lambda

Square both sides of equation:

Since v = 2 i + j - 3 k$\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$.

Step 2: Apply Identity

Since components are orthogonal, direct magnitude squared holds:

Pattern Recognition

Do not waste time explicitly projecting components v₁$\vec{v}_1$ and v₂$\vec{v}_2$ if only the sum of their squared magnitudes is requested. The scalar length matches the total vector length invariant under any orthogonal basis change.

Chapter Mix

Class 12 Mathematics: Vector Algebra