If \overrightarrow{a} is a non-zero vector, then (\overrightarrow{a} \cdot \stackrel{\^}{i}) \stackrel{\^}{i}+(\overrightarrow{a} \cdot \stackrel{\^}{j}) \stackrel{\^}{j}+(\overrightarrow{a} \cdot \stackrel{\^}{k}) \stackrel{\^}{k} equals______ OR The projection of the vector \stackrel{\^}{i}-\stackrel{\^}{j} on the vector \stackrel{\^}{i}+\stackrel{\^}{j} is______

CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

Question : 15 of 36

Marks: +1, -0

{a}↖{→}

is a non-zero vector, then

({a}↖{→} ⋅ {i}↖{\^}) {i}↖{\^}+({a}↖{→} ⋅ {j}↖{\^}) {j}↖{\^}+({a}↖{→} ⋅ {k}↖{\^}) {k}↖{\^}

equals______
OR
The projection of the vector

{i}↖{\^}-{j}↖{\^}

on the vector

{i}↖{\^}+{j}↖{\^}

is______

Solution:

Let

{a}↖{→}=a_1 {𝑖}↖{\^}+a_2 {𝑗}↖{\^}+a_3 {k}↖{\^}

Now, taking dot product of

{a}↖{→}

with

{𝑖}↖{\^}

, we get

{a}↖{→} ⋅ {𝑖}↖{\^}= (a_1 {𝑖}↖{\^}+a_2 {𝑗}↖{\^}+a_3 {k}↖{\^}) ⋅ {𝑖}↖{\^}=a_1 {𝑖}↖{\^} ⋅ {𝑖}↖{\^}+a_2 {𝑗}↖{\^} ⋅ {𝑖}↖{\^}+a_3 {k}↖{\^} ⋅ {𝑖}↖{\^}

⇒ {a}↖{→} ⋅ {𝑖}↖{\^}=a_1 {𝑖}↖{\^} ⋅ {𝑖}↖{\^}+a_2 ⋅ 0+a_3 ⋅ 0( ∵ {𝑗}↖{\^} ⋅ {𝑖}↖{\^}={k}↖{\^} ⋅ {𝑖}↖{\^}=0)

⇒ {a}↖{→} ⋅ {𝑖}↖{\^}=a_1

Similarly, taking dot product of

{a}↖{→}

with

{𝑗}↖{\^}

and

{k}↖{\^}

, we get

{a}↖{→} ⋅ {𝑗}↖{\^}=a_2 \;\text " and "\; {a}↖{→} ⋅ {k}↖{\^}=a_3

⇒({a}↖{→} ⋅ {𝑖}↖{\^}) {𝑖}↖{\^}+({a}↖{→} ⋅ {𝑗}↖{\^}) {𝑗}↖{\^}+({a}↖{→} ⋅ {k}↖{\^}) {k}↖{\^}=a_1 {𝑖}↖{\^}+a_2 {𝑗}↖{\^}+a_3 {k}↖{\^}={a}↖{→}

{a}↖{→}

is any non-zero vector, then

({a}↖{→} ⋅ {i}↖{\^}) {i}↖{\^}+({a}↖{→} ⋅ {j}↖{\^}) {j}↖{\^}+({a}↖{→} ⋅ {k}↖{\^}) {k}↖{\^}

equals

{a}↖{→}

.
OR
Let

{a}↖{→}={𝑖}↖{\^}-{𝑗}↖{\^}

and

{b}↖{→}={𝑖}↖{\^}+{𝑗}↖{\^}

.
Now, projection of vector

{a}↖{→}

and

{b}↖{→}

is given by,

\;{1}/{|{b}↖{→}|}({a}↖{→} ⋅ {b}↖{→})=\;{1}/{{√{1+1}}}\{1.1+(-1)(1)\}=\;{1}/{{√{2}}}(1-1)=0

Hence, the projection of vector

{a}↖{→}

{b}↖{→}

is 0 .

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