Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 17 of 72

Marks: +1, -0

Let

\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k},\quad \vec{b}=8\vec{i}+7\vec{j}-3\vec{k}

\vec{c}

\vec{a} \times \vec{c} = \vec{b}

\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4

|\vec{a} + \vec{c}|^2

Solution:

Concept:

We use the cross product equation to find unknown components of vector

\vec{c}

, then use the dot product condition to solve for the unknowns.

Explanation:

Let

\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}

Given

\vec{a} \times \vec{c} = \vec{b}

: compute the cross product.

\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ x & y & z \end{vmatrix} = (2z-2y)\hat{i} + (z+2x)\hat{j} + (-y-2x)\hat{k}

This equals

\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}

So we have equations:

2z - 2y = 8

z + 2x = 7

-y - 2x = -3

From the last two:

y = 3 - 2x

z = 7 - 2x

Now use

\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k}) = 4

→

x + y + z = 4

Substitute

y

and

z

x + (3-2x) + (7-2x) = 4

→

10 - 3x = 4

→

x = 2

Then

y = -1

z = 3

. So

\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}

Now

\vec{a} + \vec{c} = (-\hat{i}+2\hat{j}+2\hat{k}) + (2\hat{i}-\hat{j}+3\hat{k}) = \hat{i} + \hat{j} + 5\hat{k}

|\vec{a}+\vec{c}|^2 = 1^2 + 1^2 + 5^2 = 1+1+25 = 27

Answer:

27

(Option C)

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