If P = (a + 1)i + aj + ak, Q = ai + (a + 1) j + ak , Ř = ai + aj + (a + 1)k and P,Q,R are coplanar vectors and s(P . Q )² - y |R ×Q| = 0 , then value of y
is.
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Answer:
Given,
p
ˉ
=(a+1)
i
^
+a
j
^
+a
k
^
q
ˉ
=a
i
^
+(a+1)
j
^
+a
k
^
r
ˉ
=a
i
^
+a
j
^
+(a+1)
k
^
∣
∣
∣
∣
∣
∣
∣
∣
a+1
a
a
a
a+1
a
a
a
a+1
∣
∣
∣
∣
∣
∣
∣
∣
=0⇒a+1+a+a=0
a=
3
−1
p
ˉ
=
3
2
i
^
−
3
1
j
^
−
3
1
k
^
q
ˉ
=
3
1
(−
i
^
+2
j
^
−
k
^
)
r
ˉ
=
3
1
(−
i
^
−
j
^
+2
k
^
)
p
ˉ
.
q
ˉ
=
9
1
(−2−2+1)=
3
−1
r
ˉ
×
q
ˉ
=
9
1
∣
∣
∣
∣
∣
∣
∣
∣
i
−1
−1
j
2
−1
k
−1
2
∣
∣
∣
∣
∣
∣
∣
∣
=
9
1
(i(4−1)−j(−2−1)+k(1+2))
=
9
1
(3i+3j+3k)
=
3
i+j+k
∣
r
ˉ
×
q
ˉ
∣=
3
1
3
⇒∣
r
ˉ
×
q
ˉ
∣
2
=
3
1
⇒3(
p
ˉ
.
q
ˉ
)
2
−λ∣
r
×
q
∣
2
=0
⇒3.
9
1
−λ.
3
1
=0
⇒λ=1
∴
λ=1
...Answer
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