The vectors a, b and c are of the same length | vedclass

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(A) Let $c = (c_1, c_2, c_3).$ Since $|a| = |b| = |c|,$ we have $|c|^2 = |a|^2 = |b|^2 = 1^2 + 1^2 = 2.$ Thus,$c_1^2 + c_2^2 + c_3^2 = 2.$ Given that

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$(1, 0, 1)$

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(A) Let $c = (c_1, c_2, c_3).$ Since $|a| = |b| = |c|,$ we have $|c|^2 = |a|^2 = |b|^2 = 1^2 + 1^2 = 2.$ Thus,$c_1^2 + c_2^2 + c_3^2 = 2.$ Given that the vectors are taken pairwise at equal angles $\varphi,$ we have $\cos \varphi = \frac{a \cdot b}{|a||b|} = \frac{(i+j) \cdot (j+k)}{2} = \frac{1}{2}.$ Since the angle between $a$ and $c$ is $\varphi,$ we have $\frac{a \cdot c}{|a||c|} = \frac{1}{2} \implies \frac{c_1 + c_2}{2} = \frac{1}{2} \implies c_1 + c_2 = 1.$ Similarly,for $b$ and $c,$ $\frac{b \cdot c}{|b||c|} = \frac{1}{2} \implies \frac{c_2 + c_3}{2} = \frac{1}{2} \implies c_2 + c_3 = 1.$ From these,$c_1 = 1 - c_2$ and $c_3 = 1 - c_2.$ Substituting into $c_1^2 + c_2^2 + c_3^2 = 2,$ we get $(1 - c_2)^2 + c_2^2 + (1 - c_2)^2 = 2.$ $1 - 2c_2 + c_2^2 + c_2^2 + 1 - 2c_2 + c_2^2 = 2 \implies 3c_2^2 - 4c_2 = 0.$ This gives $c_2(3c_2 - 4) = 0,$ so $c_2 = 0$ or $c_2 = \frac{4}{3}.$ If $c_2 = 0,$ then $c_1 = 1$ and $c_3 = 1,$ giving $c = (1, 0, 1).$ If $c_2 = \frac{4}{3},$ then $c_1 = -\frac{1}{3}$ and $c_3 = -\frac{1}{3},$ giving $c = (-\frac{1}{3}, \frac{4}{3}, -\frac{1}{3}).$ Since $(1, 0, 1)$ is an option,the correct answer is $A$.

The vectors $a, b$ and $c$ are of the same length and taken pairwise,they form equal angles. If $a = i + j$ and $b = j + k,$ then the co-ordinates of $c$ are

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