Let $\vec{p}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ . If for some real numbers α, β and γ we have $15\hat{i}+10\hat{j}+6\hat{k}=α(2\vec{p}+\vec{q})+β(\vec{p}-2\vec{q})+γ(\vec{p}\times\vec{q})$ , then the value of γ is ________.

Question

Let $\vec{p}=2\hat{i}+\hat{j}+3\hat{k}$  and  $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ . If for some real numbers α, β and γ we have $15\hat{i}+10\hat{j}+6\hat{k}=α(2\vec{p}+\vec{q})+β(\vec{p}-2\vec{q})+γ(\vec{p}\times\vec{q})$ , then the value of γ is ________.

cdquestions Admin · Accepted Answer

Vector Operations  We are given: $$ \vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{q} = \hat{i} - \hat{j} + \hat{k} $$ Step 1: Calculate Linear Combinations of Vectors First, compute: $$ 2\vec{p} + \vec{q} = 2(2\hat{i} + \hat{j} + 3\hat{k}) + (\hat{i} - \hat{j} + \hat{k}) $$ Simplify: $$ 2\vec{p} + \vec{q} = 4\hat{i} + 2\hat{j} + 6\hat{k} + \hat{i} - \hat{j} + \hat{k} = 5\hat{i} + \hat{j} + 7\hat{k} $$ Next, compute: $$ \vec{p} - 2\vec{q} = (2\hat{i} + \hat{j} + 3\hat{k}) - 2(\hat{i} - \hat{j} + \hat{k}) $$ Simplify: $$ \vec{p} - 2\vec{q} = 2\hat{i} + \hat{j} + 3\hat{k} - 2\hat{i} + 2\hat{j} - 2\hat{k} = \hat{j} + \hat{k} $$ Step 2: Compute the Cross Product $\vec{p} \times \vec{q}$ Using the determinant formula for the cross product: $$ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 1 & -1 & 1 \end{vmatrix} $$ Expanding the determinant: $$ \vec{p} \times \vec{q} = \hat{i} \begin{vmatrix} 1 & 3 \ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \ 1 & -1 \end{vmatrix} $$ Simplify: $$ \vec{p} \times \vec{q} = \hat{i}(1 \cdot 1 - (-1 \cdot 3)) - \hat{j}(2 \cdot 1 - 3 \cdot 1) + \hat{k}(2 \cdot (-1) - 1 \cdot 1) $$ $$ \vec{p} \times \vec{q} = \hat{i}(1 + 3) - \hat{j}(2 - 3) + \hat{k}(-2 - 1) $$ $$ \vec{p} \times \vec{q} = 4\hat{i} + \hat{j} - 3\hat{k} $$ Step 3: Form and Solve the Equation We now solve for $\alpha$, $\beta$, and $\gamma$ such that: $$ \alpha(5\hat{i} + \hat{j} + 7\hat{k}) + \beta(\hat{j} + \hat{k}) + \gamma(4\hat{i} + \hat{j} - 3\hat{k}) = 15\hat{i} + 10\hat{j} + 6\hat{k} $$ Equating Components: - For $\hat{i}$: $$ 5\alpha + 4\gamma = 15 $$ - For $\hat{j}$: $$ \alpha + \beta + \gamma = 10 $$ - For $\hat{k}$: $$ 7\alpha + \beta - 3\gamma = 6 $$ Step 4: Solve the System of Equations From the first equation: $$ \alpha = \frac{15 - 4\gamma}{5} $$ Substitute $\alpha$ into the other two equations and solve: $$ \beta = \frac{11}{5}, \quad \gamma = 2, \quad \alpha = \frac{7}{5} $$ Final Answer: $\alpha = \frac{7}{5}, \beta = \frac{11}{5}, \gamma = 2$

List-I	List-II
(A) 4î − 2ĵ − 4k̂	(I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂
(B) 4î − 4ĵ + 2k̂	(II) Direction ratios are −2, 1, 2
(C) 2î − 4ĵ + 4k̂	(III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6)
(D) 4î − ĵ − 2k̂	(IV) Dot product with −2î + ĵ + 3k̂ is 10

Let \(\vec{p}=2\hat{i}+\hat{j}+3\hat{k}\) and \(\vec{q}=\hat{i}-\hat{j}+\hat{k}\). If for some real numbers α, β and γ we have
\(15\hat{i}+10\hat{j}+6\hat{k}=α(2\vec{p}+\vec{q})+β(\vec{p}-2\vec{q})+γ(\vec{p}\times\vec{q})\),
then the value of γ is ________.

Correct Answer: 2

Solution and Explanation

Vector Operations

Step 1: Calculate Linear Combinations of Vectors

Step 2: Compute the Cross Product \(\vec{p} \times \vec{q}\)

Step 3: Form and Solve the Equation

Equating Components:

Step 4: Solve the System of Equations

Final Answer:

Top Questions on Vector Algebra

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Let \(\vec{p}=2\hat{i}+\hat{j}+3\hat{k}\) and \(\vec{q}=\hat{i}-\hat{j}+\hat{k}\). If for some real numbers α, β and γ we have\(15\hat{i}+10\hat{j}+6\hat{k}=α(2\vec{p}+\vec{q})+β(\vec{p}-2\vec{q})+γ(\vec{p}\times\vec{q})\),then the value of γ is ________.