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Vector Algebra Part 3

Section: Mathematics

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Question : 42 of 72

Marks: +1, -0

Let

\vec{a}=2\hat{i}+\alpha\hat{j}+\hat{k}, \;\; \vec{b}=-\hat{i}+\hat{k}, \;\; \vec{c}=\beta\hat{j}-\hat{k}

\alpha

\beta

are integers and

\alpha\beta=-6

. Let the values of the ordered pair

(\alpha, \beta)

for which the area of the parallelogram of diagonals

\vec{a}+\vec{b}

\vec{b}+\vec{c}

\frac{\sqrt{21}}{2}

(\alpha_1, \beta_1)

(\alpha_2, \beta_2)

\alpha_1^2+\beta_1^2-\alpha_2\beta_2

[9 Apr 2024 Shift 2]

17
24
21
19

Solution: 👈: Video Solution

\;\;\text{Area of parallelogram}\; =\;\frac{1}{2}\left|\vec{d}_1\times\vec{d}_2\right|

\;A=\;\frac{1}{2}\left|(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})\right|=\;\frac{\sqrt{21}}{2}

\;\;\text{so,}\; \vec{a}+\vec{b}=\hat{i}+\alpha\hat{j}+2\hat{k}

\;\vec{b}+\vec{c}=-\hat{i}+\beta\hat{j}

\;(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&\alpha&2\\-1&\beta&0\end{vmatrix}

\;=\hat{i}(-2\beta)-\hat{j}(2)+\hat{k}(\beta+\alpha)

\;|(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})|=\sqrt{4\beta^2+4+(\alpha+\beta)^2}=\sqrt{21}

\;4\beta^2+4+\alpha^2+\beta^2+2\alpha\beta=21

\;\alpha^2+5\beta^2-12=17

\;\alpha^2+5\beta^2=29

\;\text{and}\; \alpha\beta=-6

and given

\alpha_{i}\beta

are integers

\;\;\text{so,}\;

\;\alpha=-3, \beta=2

or

\;\alpha=3, \;\; \beta=-2

\;(\alpha_1, \beta_1)=(-3,2)

\;(\alpha_2, \beta_2)=(3,-2)

\;\alpha_1^2+\beta_1^2-\alpha_2\beta_2=9+4+6=19

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