CBSE Class 12 Math 2026 All Sets Solved Paper

Question : 15 of 20

Marks: +1, -0

| {a}↖{→} | =8, | {b}↖{→} | =3

and

| {a}↖{→} × {b}↖{→} | =12

, then the value of

| {a}↖{→} ⋅ {b}↖{→} |

Solution:

| {a}↖{→} | =8, | {b}↖{→} | =3, | {a}↖{→} × {b}↖{→} | =12

| {a}↖{→} × {b}↖{→} | = | {a}↖{→}||{b}↖{→} | \sin θ

12=8 × 3 × \sin θ

\sin θ=\;{12}/{24}=\;{1}/{2}

Now,

\cos ^{2} θ=1-\sin ^{2} θ

=1-\;{1}/{4}

=\;{3}/{4}

∴ \cos θ= ± \;{{√{3}}}/{2}

{a}↖{→} ⋅ {b}↖{→}= | {a}↖{→}||{b}↖{→} | \cos θ

=8 × 3 × \;{{√{3}}}/{2}

=12 {√{3}}

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