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Question: Vector coplanar with vectors **i** + **j** and **j** + **k** and parallel to the vector 2**i** – 2**...

Vector coplanar with vectors i + j and j + k and parallel to the vector 2i – 2j – 4k, is

A

i – k

B

i – j – 2k

C

i + j – k

D

3i + 3j – 6k

Answer

i – j – 2k

Explanation

Solution

Let vector be ai+bj+cka\mathbf{i} + b\mathbf{j} + c\mathbf{k}.

∵ai+bj+ck,i+j,j+k\because a\mathbf{i} + b\mathbf{j} + c\mathbf{k},\mathbf{i} + \mathbf{j},\mathbf{j} + \mathbf{k} are coplanar.

∴ ∣abc110011∣=0\left| \begin{matrix} a & b & c \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} \right| = 0 ⇒a−b+c=0\Rightarrow a - b + c = 0

Also, since (ai+bj+ck)∣∣(2i−2j−4k)(a\mathbf{i} + b\mathbf{j} + c\mathbf{k})||(2\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})

∴ (ai+bj+ck)×(2i−2j−4k)=0(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}) \times (2\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}) = 0

i.e., ∣ijkabc2−2−4∣=0\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & b & c \\ 2 & - 2 & - 4 \end{matrix} \right| = 0

⇒ i(−4b+2c)−j(−4a−2c)+k(−2a−2b)=0\mathbf{i}( - 4b + 2c) - \mathbf{j}( - 4a - 2c) + \mathbf{k}( - 2a - 2b) = 0

⇒ −4b+2c=0,4a+2c=0,2a+2b=0- 4b + 2c = 0,4a + 2c = 0,2a + 2b = 0

⇒ c2=b1,\frac{c}{2} = \frac{b}{1}, c2=a−1,\frac{c}{2} = \frac{a}{- 1}, a−1=b1\frac{a}{- 1} = \frac{b}{1}

i.e., a−1=b1=c2\frac{a}{- 1} = \frac{b}{1} = \frac{c}{2} or a1=b−1=c−2\frac{a}{1} = \frac{b}{- 1} = \frac{c}{- 2}

∴ Required vector is i−j−2k.\mathbf{i} - \mathbf{j} - 2\mathbf{k}.