If \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \quad \vec{b}=3 \hat{i}+3 \hat{j}+\hat{k} \quad\) and \(\overrightarrow{\mathrm{c}}=\mathrm{c}_{1} \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \hat{\mathrm{k}}\) are coplanar vectors and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=5, \overrightarrow{\mathrm{~b}} \perp \overrightarrow{\mathrm{c}}\), then \(122\left(\mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}\right)\) is equal to \_\_\_\_ .

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Vector Algebra

a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}

and

c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}

are coplanar vectors and

a \cdot c = 5, b ⊥ c

, then

122 (c_{1} + c_{2} + c_{3})

is equal to
____ .

Answer: 150

The three scalar conditions give equations for

c_{1}, c_{2}, c_{3}

. From

a \cdot c = 5

comes

2 c_{1} + c_{2} + 3 c_{3} = 5

; from

b ⊥ c

comes

3 c_{1} + 3 c_{2} + c_{3} = 0

; and coplanarity of

a, b, c

means their scalar triple product is zero, which with

a \times b = (- 8, 7, 3)

reads

- 8 c_{1} + 7 c_{2} + 3 c_{3} = 0

. Subtracting the first equation from the coplanarity equation eliminates

c_{3}

and gives

10 c_{1} - 6 c_{2} = 5

, while eliminating

c_{3}

between the first two gives

- 7 c_{1} - 8 c_{2} = 5

. Solving this pair yields

c_{1} = \frac{10}{122}

c_{2} = - \frac{85}{122}

and

c_{3} = \frac{225}{122}

, so

c_{1} + c_{2} + c_{3} = \frac{10 - 85 + 225}{122} = \frac{150}{122}

and

122 (c_{1} + c_{2} + c_{3}) = 150

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer