Let a = i + 2 j + 3 k, b = 3 i + j - k and c be three vectors such that c is coplanar with a and b. If the vector c is perpendicular to b and a c = 5, then | c| is equal to:

Given: a = i + 2 j + 3 k, b = 3 i + j - k. Vector c is coplanar with a and b, c b, and a c = 5. Find: | c|. Since c is coplanar with a and b, write c = m a + n b Using c b, c b = 0 Substitute c = m a + n b: (m a + n b) b = 0 m( a b) + n( b b) = 0 Now calculate the dot products: a b = ( i + 2 j + 3 k) (3 i + j - k) = 3 + 2 - 3 = 2 b b = 3^2 + 1^2 + (-1)^2 = 11 Hence, 2m + 11n = 0 so m = - 112n Now use the condition a c = 5: a (m a + n b) = 5 m( a a) + n( a b) = 5 Compute a a = 1^2 + 2^2 + 3^2 = 14 Therefore, 14m + 2n = 5 Substitute m = - 112n: 14 (- 112n) + 2n = 5 -77n + 2n = 5 -75n = 5 n = - 115 Thus, m = - 112 (- 115) = 1130 Now find c: c = 1130 a - 115 b Substituting components, c = ( 1130 - 15) i + ( 1115 - 115) j + ( 3330 + 15) k c = 16 i + 23 j + 1115 k Therefore, | c| = ( 16)^2 + ( 23)^2 + ( 1115)^2 | c| = 136 + 49 + 121225 | c| = 25900 + 400900 + 484900 | c| = 909900 | c| = 116 Therefore, the magnitude of c is 116. The correct option is D.

Let a = i + 2 j + 3 k, b = 3 i + j - k and c be three vectors suc… — Solution | JEE Mathematics

MCQMediumJEE 2025Basics of Vectors

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \, \vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ . If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$ , then $|\vec{c}|$ is equal to:

A
$\frac{1}{\sqrt{3}}$
B
$18$
C
$16$
D
$\sqrt{\frac{11}{6}}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ . Vector $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ , $\vec{c} \perp \vec{b}$ , and $\vec{a} \cdot \vec{c} = 5$ .

Find: $|\vec{c}|$ .

Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ , write

\vec{c} = m\vec{a} + n\vec{b}

Using $\vec{c} \perp \vec{b}$ ,

\vec{c} \cdot \vec{b} = 0

Substitute $\vec{c} = m\vec{a} + n\vec{b}$ :

(m\vec{a} + n\vec{b}) \cdot \vec{b} = 0

m(\vec{a} \cdot \vec{b}) + n(\vec{b} \cdot \vec{b}) = 0

Now calculate the dot products:

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

\vec{b} \cdot \vec{b} = 3^2 + 1^2 + (-1)^2 = 11

Hence,

2m + 11n = 0

m = -\frac{11}{2}n

Now use the condition $\vec{a} \cdot \vec{c} = 5$ :

\vec{a} \cdot (m\vec{a} + n\vec{b}) = 5

m(\vec{a} \cdot \vec{a}) + n(\vec{a} \cdot \vec{b}) = 5

Compute

\vec{a} \cdot \vec{a} = 1^2 + 2^2 + 3^2 = 14

Therefore,

14m + 2n = 5

Substitute $m = -\frac{11}{2}n$ :

14\left(-\frac{11}{2}n\right) + 2n = 5

-77n + 2n = 5

-75n = 5

n = -\frac{1}{15}

Thus,

m = -\frac{11}{2}\left(-\frac{1}{15}\right) = \frac{11}{30}

Now find $\vec{c}$ :

\vec{c} = \frac{11}{30}\vec{a} - \frac{1}{15}\vec{b}

Substituting components,

\vec{c} = \left(\frac{11}{30} - \frac{1}{5}\right)\hat{i} + \left(\frac{11}{15} - \frac{1}{15}\right)\hat{j} + \left(\frac{33}{30} + \frac{1}{5}\right)\hat{k}

\vec{c} = \frac{1}{6}\hat{i} + \frac{2}{3}\hat{j} + \frac{11}{15}\hat{k}

Therefore,

|\vec{c}| = \sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{11}{15}\right)^2}

|\vec{c}| = \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{121}{225}}

|\vec{c}| = \sqrt{\frac{25}{900} + \frac{400}{900} + \frac{484}{900}}

|\vec{c}| = \sqrt{\frac{909}{900}}

|\vec{c}| = \sqrt{\frac{11}{6}}

Therefore, the magnitude of $\vec{c}$ is $\sqrt{\frac{11}{6}}$ . The correct option is D.

Common mistakes

Assuming coplanar means parallel. Here, coplanar only means $\vec{c}$ can be written as a linear combination of $\vec{a}$ and $\vec{b}$ , so use $\vec{c} = m\vec{a} + n\vec{b}$ instead.
Using the perpendicular condition incorrectly. Since $\vec{c} \perp \vec{b}$ , the correct equation is $\vec{c} \cdot \vec{b} = 0$ , not $\vec{a} \cdot \vec{b} = 0$ or componentwise equality.
Making sign errors while computing $\vec{a} \cdot \vec{b}$ . The term involving $-\hat{k}$ contributes $-3$ , so $\vec{a} \cdot \vec{b} = 3 + 2 - 3 = 2$ , not $8$ .

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