If {a} = {i}+{j}+{k}, {c} = {j}-{k}, {a}×{b} = {c} and {a}·{b} = 3, then {b} =

Correct Answer is : 13(5i‾+2j‾+2k‾)1/3({5i} + {2j} + {2k})31(5i+2j+2k)

Correct Answer is : 13(5i‾+2j‾+2k‾)1/3({5i} + {2j} + {2k})31(5i+2j+2k)

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TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 35 of 160

Marks: +1, -0

\overline{a}

\overline{i}+\overline{j}+\overline{k}

\overline{c}

\overline{j}-\overline{k}

\overline{a}\times\overline{b}

\overline{c}

\overline{a}\cdot\overline{b}

3

\overline{b}

Solution:

\text{Let } b = x\hat{i} + y\hat{j} + z\hat{k}

a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & l & 1 \\ x & y & z \end{vmatrix}

= \hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x)

= (z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k}

Given,

a \times b = c

\Rightarrow (z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k} = 0\cdot\hat{i} + 1\cdot\hat{j} + (-1)\hat{k}

On equating,

z-y = 0 \Rightarrow y = z

x-z = 1 \Rightarrow z = x-1

y-x = -1 \Rightarrow y = x-1

Also, given a

b = 3

\Rightarrow \;\; x+y+z = 3 \Rightarrow x+(x-1)+(x-1) = 3

\Rightarrow \;\; 3x = 5 \Rightarrow x = \frac{5}{3}

\therefore \;\; z = y = x-1 = \frac{5}{3} - 1 = \frac{2}{3}

\therefore \;\; b = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}

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