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AP EAPCET 05-Jul-2022 Shift 1 Paper

Section: Mathematics

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Question : 31 of 160

Marks: +1, -0

If

\overline{a}

is collinear with

\overline{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}

\overline{a} \cdot \overline{b}=27

\overline{a} \mid =

1
2
3
2

Solution: 👈: Video Solution

Let

a=x\hat{i}+y\hat{j}+z\hat{k}, \text{bo} b=3\hat{i}+6\hat{j}+6\hat{k}

.

a \cdot b=27

According to question,

\text{bo } a

is a collinear with

\text{bo } b

, then

a=\lambda b

....(i)

\Rightarrow \cdots \;\text{bo } a \cdot \text{bo } b=27

\Rightarrow \;\lambda \text{bo } b \cdot \text{bo } b\;=27

\Rightarrow \;\lambda |\text{bo } b|^2\;=27

[from Eq. (i)]

\;\lambda=\;\frac{27}{\left(\sqrt{(3)^2+(6)^2+(6)^2}\right)^2}. =\;\frac{27}{(9)^2}

\;\lambda=\;\frac{1}{3}

\;a=\;\frac{1}{3}(3\hat{i}+6\hat{j}+6\hat{k})\Rightarrow a=\hat{i}+2\hat{j}+2\hat{k}

So,

|a|\;=\sqrt{1+(2)^2+(2)^2}

\;=\sqrt{9}=3\;\text{units}\;

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