A vector whose modulus is sqrt{51} and makes | vedclass

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(D) Let the required vector be $d = d_1i + d_2j + d_3k$,where $d_1^2 + d_2^2 + d_3^2 = 51$ (given) .....$(i)$Since $a, b, c$ are unit vectors,the cond

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$\pm(5i - j - 5k)$

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(D) Let the required vector be $d = d_1i + d_2j + d_3k$,where $d_1^2 + d_2^2 + d_3^2 = 51$ (given) .....$(i)$Since $a, b, c$ are unit vectors,the condition that $d$ makes the same angle $	heta$ with them implies:$\cos 	heta = \frac{d \cdot a}{|d||a|} = \frac{d \cdot b}{|d||b|} = \frac{d \cdot c}{|d||c|}$Since $|d| = \sqrt{51}$ and $|a| = |b| = |c| = 1$,we have $d \cdot a = d \cdot b = d \cdot c$.Substituting the vectors:$\frac{1}{3}(d_1 - 2d_2 + 2d_3) = \frac{1}{5}(-4d_1 - 3d_3) = d_2$From $\frac{1}{3}(d_1 - 2d_2 + 2d_3) = d_2$,we get $d_1 - 2d_2 + 2d_3 = 3d_2 \Rightarrow d_1 - 5d_2 + 2d_3 = 0$.From $\frac{1}{5}(-4d_1 - 3d_3) = d_2$,we get $-4d_1 - 3d_3 = 5d_2 \Rightarrow 4d_1 + 5d_2 + 3d_3 = 0$.Solving these equations using cross-multiplication:$\frac{d_1}{(-5)(3) - (2)(5)} = \frac{d_2}{(2)(4) - (1)(3)} = \frac{d_3}{(1)(5) - (-5)(4)}$$\frac{d_1}{-15 - 10} = \frac{d_2}{8 - 3} = \frac{d_3}{5 + 20} \Rightarrow \frac{d_1}{-25} = \frac{d_2}{5} = \frac{d_3}{25}$Dividing by $-5$,we get $\frac{d_1}{5} = \frac{d_2}{-1} = \frac{d_3}{-5} = \lambda$.So,$d = \lambda(5i - j - 5k)$.Using $|d|^2 = 51$,we have $\lambda^2(25 + 1 + 25) = 51 \Rightarrow 51\lambda^2 = 51 \Rightarrow \lambda = \pm 1$.Thus,the required vector is $\pm(5i - j - 5k)$.

$A$ vector whose modulus is $\sqrt{51}$ and makes the same angle with $a = \frac{i - 2j + 2k}{3}$,$b = \frac{-4i - 3k}{5}$,and $c = j$ is:

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