Vector Algebra JEE 2022: Solving the Advanced Multi-Correct PYQ

What is the JEE Advanced 2022 Vector Algebra question?

The decisive step in Vector Algebra JEE 2022 is recognising that the skew-symmetric matrix gives a cross product in the order shown. This hard, multi-correct Mathematics question appeared in JEE Advanced 2022 Paper 2. Its expected solving time is 300 seconds, based on its hard difficulty rating and the required option-by-option checks.

The vectors are

They satisfy $b_2{b}_3>0$ and $\overrightarrow{a}·\overrightarrow{b}=0$

The skew-symmetric matrix formed from the components of vector c acts on vector b and gives vector a minus vector c:

Test these four claims:

How do you convert the conditions into a one-parameter system?

Reduce vector b to one parameter, then identify the matrix product as vector c crossed with vector b. This order keeps the Vector Algebra JEE 2022 solution within the expected 300 seconds. Beginning component calculations before identifying the cross-product order creates an avoidable sign risk.

From the scalar-product condition, $3+b_2-b_3=0$

Hence, $b_3=b_2+3$

Set $b_2=t$

Then $\overrightarrow{b}=(1,t,t+3)$

The product restriction becomes $t(t+3)>0$

Therefore,

Recognising the matrix structure is the key step. The same pattern appears in Matrices and Determinants JEE 2021: Identities and Solution.

Expanding the matrix-vector product gives

It is not vector b crossed with vector c. Reversing the order reverses every sign: $\overrightarrow{b}\times \overrightarrow{c}=-(\overrightarrow{c}\times \overrightarrow{b})$

The matrix condition is therefore $\overrightarrow{c}\times \overrightarrow{b}=\overrightarrow{a}-\overrightarrow{c}$

How do you solve for the three components of vector c?

The official component-elimination method gives all three components in terms of the same positive denominator. Write the three scalar equations, eliminate the second and third components in order, and only then substitute back. No denominator can vanish because its completed-square form is strictly positive.

Comparing components gives

$t{c}_1-c_2+c_3=-1$ Subtract the third equation from the second:

Thus,

Use the third equation: $c_3=-1-t{c}_1+c_2$

Substitute the expression for the second component:

Therefore,

Put both expressions into the first equation:

Simplification gives

Define $D=2t^2+6t+11$

Then

Substituting this into the expression for the second component gives

The third component is

Thus,

The denominator cannot vanish because

Which options are correct in the JEE Advanced 2022 question?

The official correct options are B, C and D. Option A fails because the required dot product equals a positive nonzero value. Option B follows from exact cancellation. Options C and D follow from the restriction on the parameter, not from checking selected values.

Why is option A false?

The dot product is nonzero, so vector a is not perpendicular to vector c. Substitute the three components carefully, especially the negative third component.

Therefore,

Since $D>0$ the dot product is nonzero. Option A is false.

Why is option B true?

The dot product of vectors b and c cancels exactly for every allowed value of the parameter. No inequality is needed for this option.

Substitution gives

Thus vector b is perpendicular to vector c. Option B is true.

Why is option C true?

The squared magnitude of vector b exceeds 10 because the given parameter restriction makes the difference positive. Compare squared magnitudes to avoid unnecessary square roots.

Therefore,

Hence, $|\overrightarrow{b}|>\sqrt{10}$

Option C is true.

Why is option D true?

The parameter restriction proves the stronger result that the magnitude of vector c is less than 1. This immediately settles the stated upper bound.

Use the identity

Since $t(t+3)>0$ and $D>0$ the right side is positive. Therefore, $|\overrightarrow{c}{|}^2<1$

The stronger conclusion is $|\overrightarrow{c}|<1$

This immediately implies $|\overrightarrow{c}|\le \sqrt{11}$

Option D is true. The official correct options are B, C and D.

How can a sign error incorrectly produce option A?

The error occurs while evaluating the final term of the dot product. The third component is negative, so subtracting it produces a positive term. Cancelling the parameter terms without retaining the constant term incorrectly makes the full numerator appear to be zero.

Here,

Therefore,

The complete numerator is $-6t+(3t+11)+3t=11$

The parameter terms cancel, but the constant 11 remains. Cancelling the entire numerator after seeing the parameter terms disappear would incorrectly give $\overrightarrow{a}·\overrightarrow{c}=0$

That error leads to marking option A. Keep the common denominator and every constant term until the numerator has been fully combined.

Which related Vector Algebra questions should you practise?

Practise one question each on scalar-product parameterisation, skew-symmetric matrix recognition and norm comparison. These are the three operations used in this PYQ. Solve them by direct component work, since that is the reliable method under the expected 300-second limit.

Can you solve this scalar-product parameter question?

The parameter is fixed by the scalar-product condition, after which the norm comparison is immediate.

Given

and $\overrightarrow{p}·\overrightarrow{q}=0$ find the parameter and test whether the magnitude of vector q exceeds 2.

Answer checkpoint:

Then $|\overrightarrow{q}|=\sqrt{5}>2$

Does this matrix represent vector x crossed with vector y or the reverse?

The matrix represents vector x crossed with vector y. Verify the order by expanding all three components rather than identifying it from memory.

For

identify the product represented when

acts on vector y.

Answer checkpoint: Component expansion gives

Hence it represents $\overrightarrow{x}\times \overrightarrow{y}$

Can you use a parameter restriction to prove a norm bound?

The component equations reduce the magnitude to a rational expression whose denominator is controlled by the sign restriction.

Let

with $r(r+2)>0$

Given $\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{w}-\overrightarrow{u}$ form the three component equations and prove

Checking cue: For

the component equations are

$r{u}_1-u_2+u_3=0$ Elimination gives

The restriction implies $(r+1{)}^2>1$ so the stated upper bound follows. For more drills of the same type, use the free past-paper archive to search JEE Advanced Mathematics questions by chapter and compare your component work with the worked solutions.