Solveeit Logo
Login

Question

Question: What vector when added to \((2 \hat{i} – 2 \hat{j} + \hat{k})\) and \((2 \hat{i} - \hat{k})\) will g...

What vector when added to (2i^–2j^+k^)(2 \hat{i} – 2 \hat{j} + \hat{k}) and (2i^−k^)(2 \hat{i} - \hat{k}) will give a unit vector along negative y-axis?

Explanation

Solution

We have given two vectors. We have to find another vector which when added to the sum of given vectors then it will give a unit vector along the negative y-axis. So, we will find out that vector after comparing the coefficients of the basis vector.

Complete step-by-step solution:
Let A⃗=(2i^–2j^+k^)\vec{A} = (2 \hat{i} – 2 \hat{j} + \hat{k}) and B⃗=(2i^−k^)\vec{B} = (2 \hat{i} - \hat{k}).
We need to find a vector C⃗\vec{C} which when added to A⃗+B⃗\vec{A} + \vec{B} gives a unit vector along the negative y-axis.
So, we first take the sum of vectors A⃗\vec{A} and B⃗\vec{B}.
A⃗+B⃗=(2i^–2j^+k^)+(2i^−k^)\vec{A} + \vec{B} = (2 \hat{i} – 2 \hat{j} + \hat{k}) + (2 \hat{i} - \hat{k})
We get,
A⃗+B⃗=4i^–2j^\vec{A} + \vec{B} = 4 \hat{i} – 2 \hat{j}
Let C⃗=(xi^+yj^+zk^)\vec{C} = (x \hat{i} + y \hat{j} + z\hat{k})
Now add vector C⃗\vec{C} to A⃗+B⃗\vec{A} + \vec{B}
A⃗+B⃗+C⃗=(4i^–2j^)+(xi^+yj^+zk^)\vec{A} + \vec{B}+ \vec{C} = ( 4 \hat{i} – 2 \hat{j}) + (x \hat{i} + y \hat{j} + z\hat{k})
Now according to the question,
(4i^–2j^)+(xi^+yj^+zk^)=−j^( 4 \hat{i} – 2 \hat{j}) + (x \hat{i} + y \hat{j} + z\hat{k}) = -\hat{j}
(4+x)i^+(y–2)j^+zk^=−j^( 4+x) \hat{i} +(y– 2) \hat{j} + z\hat{k} = -\hat{j}
Compare the coefficient of i, j and k.
(4+x)=0  ⟹  x=−4(4 + x) = 0 \implies x =-4
y−2=−1  ⟹  y=1y -2= -1 \implies y =1
z=0z = 0
Now vector C⃗\vec{C} will be,
C⃗=−4i⃗+j^\vec{C} = -4 \vec{i} + \hat{j}

Note: A vector contains both a magnitude and a direction. Geometrically, we can imagine a vector as a pointed line segment, whose length is the vector's magnitude, and with an arrow showing the direction. The direction of the vector is from its end to its head.