Let {a}=2 {i}+3 {j}+{k}, {b}=4 {i}+{j}, {c}={i}-3 {j}-7 {k}. If {r}=x {i}+y {j}+z {k}, {r} · {a}=9, {r} {b}=7, {r} · {c}=6 then (x, y, z)=

Correct Answer is : (1,3,−2)(1,3,-2)(1,3,−2)

Correct Answer is : (1,3,−2)(1,3,-2)(1,3,−2)

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AP EAPCET 06-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 31 of 160

Marks: +1, -0

Let

\overline{a}=2 \overline{i}+3 \overline{j}+\overline{k}, \overline{b}=4 \overline{i}+\overline{j}, \overline{c}=\overline{i}-3 \overline{j}-7 \overline{k}

\overline{r}=x \overline{i}+y \overline{j}+z \overline{k}, \overline{r} \cdot \overline{a}=9, \overline{r} \overline{b}=7, \overline{r} \cdot \overline{c}=6

(x, y, z)=

Solution: 👈: Video Solution

\mathbf{r} \cdot \mathbf{a}=9

\Rightarrow \;\; 2x+3y+z=9

.....(i)

\Rightarrow \mathbf{r} \cdot \mathbf{b}=7

\Rightarrow 4x+y=7

....(ii)

\Rightarrow \;\; r \cdot c=6

\Rightarrow \;\; x-3y-7z=6

.....(iii)

On adding Eqs. (i) and (iii), we get

\Rightarrow\; 3x-6z=15

\Rightarrow\; x-2z=5 \Rightarrow 2z=x-5

\Rightarrow\; z=\; \frac{1}{2}(x-5)

From Eq. (ii), we get

y=7-4x

put

z=\; \frac{1}{2}(x-5)

and

y=7-4x

in Eq. (i)

\; 2x+3y+z=9

\Rightarrow 2x+3(7-4x)+\; \frac{1}{2}(x-5)=9

\Rightarrow 4x+42-24x+x-5=18

\Rightarrow -19x+19=0 \Rightarrow x=1, y=7-4=3

\; \text{ and } \;\; z=\; \frac{1}{2}(1-5)=-2

\therefore \; (x, y, z)=(1,3,-2)

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