Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0$, $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and projection of $\vec{c}$ on $\vec{a}$ is 1, then the projection of $\vec{c}$ on $\vec{b}$ equals:

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Vector Algebra

Let

a = 4 \hat{i} + 3 \hat{j}

and

b = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}

and

c

is a vector such that

c \cdot (a \times b) + 25 = 0

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

and projection of

c

a

is 1, then the projection of

c

b

equals:

Answer: (A)

With

a \times b = 15 \hat{i} - 20 \hat{j} - 25 \hat{k}

, write

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

. The conditions become

3 x - 4 y - 5 z = - 5

x + y + z = 4

, and

\frac{4 x + 3 y}{5} = 1

, i.e.

4 x + 3 y = 5

. Solving these three linear equations gives

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

. The projection of

c

b

\frac{c \cdot b}{∣ b ∣} = \frac{6 + 4 + 15}{50} = \frac{25}{5 2} = \frac{5}{2}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option