MCQ
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 $c = j,$ will be
- A$5i + 5j + k$
- B$5i + j - 5k$
- C$5i + j + 5k$
- ✓$ \pm \,(5i - j - 5k)$
Now, each of the given vectors $a,\,\,b,\,\,c$ is a unit vector
$\cos \theta = \frac{{d\,.\,a}}{{|d|\,|a|}} = \frac{{d\,.\,b}}{{|d|\,|b|}} = \frac{{d\,.\,c}}{{|d|\,|c|}}$
or $d\,.\,a = d\,.\,b = d\,.\,c$
$|d| = \sqrt {51} $ cancels out and $|a|\,\, = \,\,|b|\, = \,|c|\, = 1$
Hence, $\frac{1}{3}({d_1} - 2{d_2} + 2{d_3}) = \frac{1}{5}( - 4{d_1} + 0{d_2} - 3{d_3}) = {d_2}$
$ \Rightarrow {d_1} - 5{d_2} + 2{d_3} = 0$ and $4{d_1} + 5{d_2} + 3{d_3} = 0$
On solving, $\frac{{{d_1}}}{5} = \frac{{{d_2}}}{{ - 1}} = \frac{{{d_3}}}{{ - 5}} = \lambda $(say)
Putting ${d_1},\,{d_2}$ and ${d_3}$ in $(i),$ we get $\lambda = \pm 1$
Hence the required vectors are $ \pm (5i - j - 5k).$
Trick : Check it with the options.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
($A$) The function $f$ is discontinuous exactly at one point in $(0,1)$
($B$) There is exactly one point in $(0,1)$ at which the function $f$ is continuous but $NOT$ differentiable
($C$) The function $\mathrm{f}$ is $NOT$ differentiable at more than three points in $(0,1)$
($D$) The minimum value of the function $f$ is $-\frac{1}{512}$