MCQ
Vector coplanar with vectors $i + j $ and $ j + k$  and parallel to the vector $2i -2j -4k$ , is
  • A
    $i -k$
  • $i -j -2k$
  • C
    $i + j -k$
  • D
    $3i + 3j -6k$

Answer

Correct option: B.
$i -j -2k$
b
(b) Let vector be $ai + bj + ck$.

$\because ai + bj + ck\,,\,i + j\,,\,\,j + k$ are coplanar.

$\therefore \left| {\,\begin{array}{*{20}{c}}a&b&c\\1&1&0\\0&1&1\end{array}\,} \right| = 0$ $ \Rightarrow a - b + c = 0$

Also, since $(ai + bj + ck\,)|\,|\,\,(2i - 2j - 4k)$

 $\therefore (ai + bj + ck) \times (2i - 2j - 4k) = 0$

i.e., $\left| {\,\begin{array}{*{20}{c}}i&j&k\\a&b&c\\2&{ - 2\,}&{ - \,4\,\,}\end{array}\,} \right| = 0$

==> $i( - \,4b + 2c) - j( - \,4a - 2c) + k( - \,2a - 2b) = 0$

==> $ - \,4b + 2c = 0,\,\,4a + 2c = 0,\,\,\,2a + 2b = 0$

==> $\frac{c}{2} = \frac{b}{1},$ $\frac{c}{2} = \frac{a}{{ - 1}},$ $\frac{a}{{ - 1}} = \frac{b}{1}$

i.e., $\frac{a}{{ - 1}} = \frac{b}{1} = \frac{c}{2}$ or $\frac{a}{1} = \frac{b}{{ - 1}} = \frac{c}{{ - 2}}$

$\therefore $ Required vector is $i - j - 2k.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The feasible region of a linear programming problem is bounded. The corresponding objective function is $Z=6 x-7 y$.
The objective function attains $\qquad$ in the feasible region.
A tetrahedron has vertices at $O(0,\,0,\,0)$, $A(1,\,2,\,1),B(2,\,1,\,3)$ and $C( - 1,\,1,\,2)$. Then the angle between the faces $OAB$ and $ABC$will be
Three points whose position vectors are $a + b,\,\,a - b$ and $a + kb$ will be collinear, if the value of   $k $ is
If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{4\text{x}^3}{1-\text{x}^4}$
  2. $-\frac{4\text{x}}{1-\text{x}^4}$
  3. $\frac{1}{4-\text{x}^4}$
  4. $-\frac{4\text{x}^3}{1-\text{x}^4}$
The value of $\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$ is:
  1. $1+\log\text{x}$
  2. $\text{x}+\log\text{x}$
  3. $\text{x}\log\text{x}(1+\log\text{x})$
  4. $\log(1+\log\text{x})$
The sum of the terms of an infinitely decreasing geometric progression is equal to the greatest value of the function $f (x) = x^3 + 3x -9$ on the interval $[- 2, 3]$ . If the difference between the first and the second term of the progression is equal to $f ' (0)$ then the common ratio of the $G.P$. is
Suppose  $f $ is such that $f( - x) = - f(x)$ for every real $x$ and $\int_{\,0}^{\,1} {f(x)\,dx = 5,} $ then $\int_{\, - \,1}^{\,0} {f(t)\,dt = } $
The solution set of the inequality $3 x+5 y<4$ is
Let $\vec a = 2\hat i + \hat j - 2\hat k,\,\vec b = \hat i + \hat j$ . If $\vec c$ is a vector such that $\vec a.\vec c + 2\left| {\vec c} \right| = 0$ and $\left| {\vec c - \vec a} \right| = \sqrt {14} $ and angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$ , then $\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right|$ is
Which of the following is not a convex set?